To prove that H(n) = n!(1/1! + 1/2! + 1/3! + ... + 1/n!) for all n ≥ 1 using induction, we need to follow the steps you've outlined.
Base Case:
For n = 1, we have H(1) = 1·H(0) + 1. Plugging in H(0) = 0 and simplifying, we get H(1) = 1·0 + 1 = 1. On the other hand, f(1) = 1!(1/1!) = 1(1) = 1. The base case holds true.
Inductive Hypothesis:
Assume that for some k > 1, H(k-1) = (1/1! + 1/2! + 1/3! + ... + 1/(k-1)!). This is our inductive hypothesis.
Inductive Step:
Using the recurrence relation, we have H(k) = k·H(k-1) + 1. Plugging in our inductive hypothesis, we get:
H(k) = k(1/1! + 1/2! + 1/3! + ... + 1/(k-1)!) + 1.
To simplify further, we can write k as k!/k!:
H(k) = k!/k! (1/1! + 1/2! + 1/3! + ... + 1/(k-1)!) + 1.
Rearranging the terms, we get:
H(k) = (1/1! + 1/2! + 1/3! + ... + 1/(k-1)!) + k!/k!.
This expression is equal to f(k), which is n!(1/1! + 1/2! + 1/3! + ... + 1/n!). Therefore, we have shown that H(k) = f(k) for the inductive step.
By induction, we have proved that H(n) = n!(1/1! + 1/2! + 1/3! + ... + 1/n!) for all n ≥ 1.
Note: It's important to clarify that H(0) should be explicitly defined as H(0) = 0 in the recurrence relation to ensure that the base case is consistent.
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Drag each tile to the correct box.
Graph the functions as transformations of f(x) = x^(2).
Arrange the parabolas with respect to the position of their vertices from left to right.
Answer:
Step-by-step explanation:
You use generally accepted facts (such as: there are approximately 6 billion people on the planet. ) without citing them is it plagiarism?
No, this not plagiarism .You can use generally accepted facts without citing them.
What defines self-plagiarism?
Self-plagiarism is defined as a type of plagiarism in which the writer republishes a work in its entirety or reuses portions of a previously written text while authoring a new work.
What is considered general knowledge in plagiarism?
Common knowledge is typically defined as facts that are unsupported by at least five reliable sources. In the discipline of composition studies, for instance, the statement "Writing is difficult" is taken for granted because at least five reliable sources may support it.How to Avoid plagiarism common knowledge?
Quoting correctly, paraphrasing and summarizing correctly, and documenting (citing) correctly are the keys to providing credit and avoiding plagiarism; you can go to page 3 of this handout for more information. Almost all of the information you discover in outside sources needs to be documented.Learn more about plagiarism
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You want to prove that FHG is similar to RXS by the SSS Similarity Theorem. Complete the proportion that is needed to use this theorem.
Answer:
[tex]\frac{FH}{RX} =\frac{HG}{XS} =\frac{FG}{RS}[/tex]
Step-by-step explanation:
The order of the letters matters in any similarity theorem.
When triangle FHG is similar to triangle RXS, the lengths of FH must be similar to RX.
When 2 similar lines are divided, that number is the factor by which one triangle can be dilated to match the other.
That being said, each of these proportion statements can be read as individual division expressions that equal the next division expression.
This means [tex]FH\div RX = HG\div XS = FG\div RS[/tex].
If you look at it this way, you can see that the sides you need to fill in must be the corresponding similar sides, so that you end up with the factor for the dilation of the 2 triangles.
Find the number of 4 digit numbers that contain at least 3 even digits
Answer:
2625 different numbers
Step-by-step explanation:
General outlineCombinatoricsSpecial circumstancesPartitioning the situationConclusionPart 1. CombinatoricsThis type of problem is classified as a "Combinatorics" problem: Counting up the number of ways something can happen.
Many times when attempting to count a large number of items, patterns are recognized and shortcuts to count are developed so that one doesn't need to physically count all of the items. Perhaps the trickiest part of counting with these shortcuts is to ensure that one does not over-count (counting a scenario more than once), although it is equally important that we don't under-count (fail to count a situation that applies).
Part 2. Special circumstancesIt should be noted that for a number to be a 4 digit number, it must have 4 digits, and thus the first digit must not be zero (despite that that would be an even digit, the number itself would not be a 4 digit number).
Part 3. Partitioning the situationThere are 2 main scenarios:
1. All 4 digits are even
2. Exactly 3 digits are even (meaning that exactly one digit is odd).
There are two sub-cases to Scenario 2:
2.1. The first digit is odd, and all three of the other digits are even.
2.2. The first digit is even, and exactly two of the other digits are even (meaning that exactly one of the last three digits is odd).
None of scenarios 1, 2.1, and 2.2 overlap, so we're not over-counting:
If all 4 digits are even, then there can't be exactly 3 even digits. If the first digit is odd, then not all 4 digits are even nor is the first digit even. If exactly two of the last three digits are even, then not all 4 digits are even, nor are all three of the last digits even.Further, this is all of the possibilities for a 4 digit number with least three even digits, so we're not under-counting.
Scenario 1 -- All 4 digits are even.
If all 4 digits are even, then the first digit has fours choices (2,4,6,8), and the next 3 digits each have 5 choices (2,4,6,8,0).
[tex]4*5*5*5=500\text{ choices}[/tex]
Scenario 2.1 -- The first digit is odd, and all three of the other digits are even
If the first digit is odd, there are 5 choices for the first digit (1,3,5,7,9), and the next 3 digits each have 5 choices (2,4,6,8,0).
[tex]5*5*5*5=625\text{ choices}[/tex]
Scenario 2.2 -- The first digit is even, and exactly two of the other digits are even
If the first digit is even, there are 4 choices for the first digit (2,4,6,8), and if exactly two of the next 3 digits are even, then there are 5 choices for each of the two even digits (2,4,6,8,0), and 5 choices for the odd digit (1,3,5,7,9), and there are "3 permuted by 2" ways of ordering those three digits.
[tex]4* \left [(5*5*5) * {}_3 \! P_2 \right ]=\\\\=4*\left [5*5*5 * \dfrac{3!}{2!} \right ]\\\\=4*\left [ 5*5*5 * \dfrac{3*2*1}{2*1} \right ] \\\\=4*\left [5*5*5 * 3 \right ] \\\\=1500\text{ choices}[/tex]
Scenario 2.2 broken down (calculated without using the permutation operation) -- The first digit is even, and exactly two of the other digits are even
Then either the second digit is odd (scenario 2.2.1), the third digit is odd (scenario 2.2.2), or the fourth digit is odd (scenario 2.2.3).
Scenario 2.2.1. The second digit is odd.
The first digit is even: 4 choices for the first digit (2,4,6,8)
The second digit is odd: 5 choices for the odd digit (1,3,5,7,9)
The third and fourth digits are even: 5 choices for each even digit (2,4,6,8,0).
[tex]4*5*5*5=500\text{ choices}[/tex]
Scenario 2.2.2. The third digit is odd.
The first digit is even: 4 choices for the first digit (2,4,6,8)
The second and fourth digits are even: 5 choices for each even digit (2,4,6,8,0).
The third digit is odd: 5 choices for the odd digit (1,3,5,7,9)
[tex]4*5*5*5=500\text{ choices}[/tex]
Scenario 2.2.3. The last digit is odd.
The first digit is even: 4 choices for the first digit (2,4,6,8)
The second and third digits are even: 5 choices for each even digit (2,4,6,8,0).
The last digit is odd: 5 choices for the odd digit (1,3,5,7,9)
[tex]4*5*5*5=500\text{ choices}[/tex]
Scenarios 2.2.1, 2.2.2, and 2.2.3 comprise all of Scenario 2.2, so [tex]500+500+500=1500\text{ choices}[/tex]
Part 4. ConclusionHow many ways can a 4 digit number be formed where at least 4 of the digits are even?
This is the sum of the choices from Scenario 1, Scenario 2.1, and Scenario 2.2, so [tex]500+625+1500=2625\text{ choices}[/tex].
The functions f(x) and g(x) are graphed.
On a coordinate plane, a curved red line with an upward arc, labeled g of x, crosses the y-axis at (0, 4) and the x-axis at (2, 0). A straight blue line with a negative slope, labeled f of x, crosses the y-axis at (0, 4) and the x-axis at (2, 0).
Which represents where f(x) = g(x)?
f(0) = g(0) and f(2) = g(2)
f(2) = g(0) and f(0) = g(4)
f(2) = g(0) and f(4) = g(2)
f(2) = g(4) and f(1) = g(1)
The function equality that represents the point where f(x) = g(x) is; A: f(0) = g(0) and f(2) = g(2)
How to interpret Graphed Functions?From the question, the curved red line represents g(x) while the straight blue line represents f(x).
Now, the equality of functions f(x) = g(x) is represented as a common function between their curves. Thus, we will just need to find a common point for both. This is;
f(x) has points (0, 4) and (2, 0).
g(x) has points (0,4) and (2,0).
It is pertinent to note that both functions have the same y-value for x=0 and x = 2. Thus, we can say that;
f(2) = g(2) and f(0) = g(0)
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Suppose you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to pull out one marble, record its color, remove it from the box and draw another marble. What is the probability of pulling out a red marble followed by a blue marble? The multiplication rule says to use P(red) P(blue).
Describe the probability of finding a red marble?
Describe the probability of finding a blue marble?
Describe the process of finding the probability of finding a red marble followed by a blue marble if the first marble was permanently removed?
What affect did removing the first marble from the box have on the problem?
Describe the probability of finding a red marble followed by the blue marble if the first marble is removed?
The probability of randomly getting first a red marble and then a blue marble is:
P = 0.083
How to find the probability?
There are:
3 blue marbles2 red marbles4 yellow marbles.For a total of 9 marbles.
The probability of getting a red marble is equal to the quotient between the number of red marbles and the total, so:
P(red) = 2/9
Then the probability of getting a blue marble is equal to the quotient between the number of blue marbles and the total, but because we already took one marble, now the total is 8.
P(blue) = 3/8
The joint probability is given by the product:
P = (2/9)*(3/8) = 0.083
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help please much appreciated if you attempt to help
196 36 correct answer is 232
Answer:
728 cm²
Step-by-step explanation:
the surface area is the area of the 6 faces.
the opposite rectangular faces are congruent , then
SA = front/ back + sides + top/ bottom
= 2(14 × 14) + 2(6 × 14) + 2(14 × 6)
= 2 × 196 + 2 × 84 + 2 × 84
= 392 + 168 + 168
= 728 cm²
7 cm
D
What is the sum of the areas of circle C and circle D?
O7ft units²
O 14 units²
O 49 units²
O98 units²
The sum of areas of circle C and circle D exists 98 square units.
How to estimate the sum of the areas of circle C and circle D?Given: Radius of circle = 7 units
Area of circle [tex]=\pi r^2[/tex] ..............(1)
Where, r be the radius of circle
The radius of circle C, r = 7 units
(1) becomes
Area of circle, C [tex]= \pi 7^2[/tex] square units
[tex]= 49\pi[/tex] units²
Area of circle, D [tex]= \pi 7^2[/tex] square units
[tex]= 49\pi[/tex] units²
Sum of areas of circles C and D,
= Area of circle C + Area of circle D
= 49 + 49
= 98 units²
The sum of areas of circles C and D exist 98 units².
Therefore, the correct answer is option d) 98 units².
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The population of a bacteria culture are tripling. after 6 days, the population has quadrupled. what is the tripling time? find the answer to the nearest thousandth. b) how long did it take for the population to double in size?
The tripling time of population is 6 days and the population will be doubled in 3.5 days.
Given that the population of a bacteria is tripling in 6 days.
We have to find the tripling time of population and the time when the population will be doubled.
Suppose the population of bacteria in beginning be x.
The population of bacteria seems like arithmetic progression in which a =x, nth term =3x and n=6.
Aritmetic progression is a sequence which is having common difference.
nth term=a+(n-1)d
3x=x+(6-1)*d
3x-x=5d
2x=5d
d=2x/5
d=0.4x
So now we have to find the value of n where population will be 2x.
nth term=a+(n-1)d
put nth term=2x, a=x,d=0.4x
2x=x+(n-1)*0.4x
2x-x=(n-1)*0.4x
x/0.4x=n-1
2.5=n-1
n=2.5+1
n=3.5
Hence the population of bacteria willbe doubled after 3.5 days.
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Expand the expression / brackets: 2y (3y + 9)=
Answer:
6y^2 + 18y
Step-by-step explanation:
2y (3y + 9)=
6y^2 + 18y
[2y (3y becomes 6y^2]
[2x( + 9) becomes 18y]
Add them to obtain 6y^2 + 18y
Select the correct answer. which equation could possibly represent the graphed function?
The function for the given graph is f(x) = (x - 4)(x + 2)(x + 4).
How to represent a graph by an equation?A graph can be represented by the equation y = f(x).In the equation, the variable x is called the independent variable and the variable y is called the dependent variable.Calculation:In the given graph,
The points that lie on the x-axis are (-4, 0), (-2, 0), and (4, 0) where y-coordinate is 0 and the x coordinates are -4, -2, and 4.
On solving the given functions,
Option A:
f(x) = (x - 4)(x + 2)(x + 4)
Since y = 0 consider f(x) becomes 0
So,
(x - 4)(x + 2)(x + 4) = 0
According to the zero product rule,
x - 4 = 0, x + 2 = 0, and x + 4 = 0
⇒ x = 4, x = -2 and x = -4.
Thus, option A represents the function of the given graph.
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Disclaimer: The question given on the portal was incomplete. Here is the complete question.
Question:
Select the correct answer. which equation could represent the graphed function (given in the figure below)?
What is the simplified form of startroot startfraction 2,160 x superscript 8 baseline over 60 x squared endfraction endroot? assume x ≠ 0.
The simplified form is [tex]6x^{3}[/tex].
What are indices in maths?
An index, or a power, is the small floating number that goes next to a number or letter.The plural of index is indices. Indices show how many times a number or letter has been multiplied by itself.The equation is -
[tex]\sqrt{\frac{2160x^{3} }{60x^{2} } }[/tex]
We can rewrite this using property of multiplication to obtain
[tex]\sqrt{\frac{2160}{60} * \frac{x^{8} }{x^{2} } }[/tex]
We simplify variable expression in the radicand using
[tex]\frac{a^{m} }{a^{n} } = a^{m - n}[/tex]
[tex]\sqrt{36 * x^{8 - 2} }[/tex]
[tex]\sqrt{36 * x^{6} }[/tex]
[tex]\sqrt{36 * (x^{3})^{2} }[/tex]
[tex]\sqrt{36} * \sqrt{(x^{3})^{2} }[/tex]
[tex]6 * x^{3}[/tex]
Therefore, the simplified form is [tex]6x^{3}[/tex]
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A parabola, with its vertex at the origin, has a directrix at y = 3. which statements about the parabola are true? select two options.
The correct options about parabola are:
The focus is located at (0,–3).
The parabola can be represented by the equation x^2 = –12y.
According to the statement
we have given that the vertex is at the origin and directrix at y = 3.
and from these given information and we have to find the all abot the parabola like its focus point etc.
So,
We know that the equation of parabola is
(x-h)^2 = 4p(y-k)
Here The vertex is (h,k). and the focus is at (h,k+p). and the directrix is y(k - p.)
So, From the given information
Vertex at the origin means that h=0 and k=0
Directrix at y = 3 means that p=-3
Directrix at the y-axis means the parabola opens upwards.
Thus, the focus is: (0,-3)
And The p-value becomes
: 4(-3) = -12.
And from all these the equation of the parabola is becomes
(x)^2 = -12y
So, The correct options about parabola are:
The focus is located at (0,–3).
The parabola can be represented by the equation x^2 = –12y.
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Question:
A parabola, with its vertex at the origin, has a directrix at y = 3. Which statements about the parabola are true? Select two options.
The focus is located at (0,–3).
The parabola opens to the left.
The p value can be determined by computing 4(3).
The parabola can be represented by the equation x2 = –12y.
The parabola can be represented by the equation y2 = 12x.
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What is the area of the trapezoid above?
A. 188 cm²
B. 196 cm²
C. 392 cm²
D. 5,236 cm²
Answer:
B. 196cm²Step-by-step explanation:
Area = h/2 (b_1+b_2)
[tex]\sf\cfrac{7}{2}\:(22+34)[/tex]
Add 22 and 34 = 56
[tex]\sf \cfrac{7}{2} \times 56[/tex]
Let's express 7/2 * 56 as a single fraction:-
[tex]\sf \cfrac{7\times 56}{2}[/tex]
Multiply 7 and 56 = 392
[tex]\sf \cfrac{392}{2}[/tex]
Divide
[tex]\sf 196[/tex]
Therefore, the area of the trapezoid is 196cm².
_______________________
what inequality is shown on the graph ?
The inequality of the graph that is given is: y < 2x - 3.
How to Write the Inequality of a Graph?First find the slope (m) = rise / run of the line.
Rise = 2 units
Run = 1 unit
Slope (m) = 2/1 = 2
Y-intercept (b) = -3.
Since the shaded side is at the left, we are going to use the ">" sign. Substitute m = 2 and b = -3 into y < mx + b.
y < 2x - 3
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A flat rectangular piece of aluminum has a perimeter of 60 inches. The length is 12 inches longer than the width. Find the width.
Answer:
width = 9 inches
Step-by-step explanation:
let width be w then length is w + 12
the perimeter ( P) is calculated as
P = 2( width + length )
here P = 60 , then
2(w + w + 12) = 60 ( divide both sides by 2 )
2w + 12 = 30 ( subtract 12 from both sides )
2w = 18 ( divide both sides by 2 )
w = 9
that is the width of the rectangle is 9 inches
Yolanda paid for her movie ticket using 28 coins, all nickels and quarters. the ticket cost $4. which system of linear equations can be used to find the number of nickels, n, and the number of quarters, q, yolanda used? n q = 28 0.05n 0.25q = 4 n q = 4 5n 25q = 28 n q = 28 5n 25q = 4 n q = 4 0.05n 0.25q = 28
The system of linear equations can be used to find the number of nickels, n, and the number of quarters, q, yolanda used is n + q = 28
n + q = 280.05n + 0.25q = 4
Simultaneous equationn + q = 28
0.05n + 0.25q = 4
n = 28 - q
substitute n = 28 - q0.05n + 0.25q = 4
0.05(28 - q) + 0.25q = 4
1.4 - 0.05q + 0.25q = 4
- 0.05q + 0.25q = 4 - 1.4
0.20q = 2.6
q = 2.6 / 0.20
q = 13
Substitute into
n + q = 28
n + 13 = 28
n = 28 - 13
n = 15
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Larry has 6 shirts and 4 pairs of jeans. How many different combinations of one shirt and one pair of jeans can Larry make
Answer:
24 combinations
Step-by-step explanation:
To get to this answer, I multiplied 6 x 4, since there are 6 shirts and 4 pairs of jeans, there will be a total of 24 combinations. For combinations of two different things, you can just multiply the amount of each item by the other item. hope this helps :)
Scores for a civil service exam are normally distributed with a mean of 75 and a standard deviation of 6.5. what score marks the difference between the bottom 10% and the top 90%?
The lowest score you can earn and still be eligible for employment is 85.6925.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
Top 5%.
At least the 100-5 = 95th percentile.
The 95th percentile is X when Z has a pvalue of 0.95. So X when Z = 1.645.
x-75 = 1.645*6.5
x = 85.692
Thus the lowest score you can earn and still be eligible for employment is 85.6925.
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Transform the given equation into a system of first order equations. (Let x1 = u, x2 = u', x3 = u'', and x4 = u'''. Enter your answers in terms of x1, x2, x3, and x4.) u(4) − u = 0
a) x1' =
b) x2' =
c) x3' =
d) x4' =
Given the 4th order linear ODE
[tex]u^{(4)} - u = 0[/tex]
we substitute
[tex]x_1 = u[/tex]
[tex]x_2 = {x_1}' = u'[/tex]
[tex]x_3 = {x_2}' = {x_1}'' = u''[/tex]
[tex]x_4 = {x_3}' = {x_2}'' = {x_1}''' = u'''[/tex]
Then the given equation transforms to
[tex]{x_4}' - x_1 = 0[/tex]
but we also need to relate this to the other derivative substitutions. This gives the system of differential equations
[tex]\begin{cases} {x_1}' = x_2 \\ {x_2}' = x_3 \\ {x_3}' = x_4 \\ {x_4}' = x_1 \end{cases}[/tex]
In matrix form,
[tex]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}' = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}[/tex]
A man mows his 100 ft by 200 ft rectangular lawn in a spiral pattern starting from the outside edge. After a bit of hard work he stops for a water break, he is 40.5% done. How wide of a strip has he mowed around the outside edge
Based on the length and width of the rectangular lawn, and the percentage the man has mowed, the strip width would be 40.5 m.
what is the width of the strip mowed already?first, find the area of the lawn:
= 100 x 200
= 20,000 ft²
the area mowed is:
= 40.5% x 20,000
= 8,100 ft²
the width is therefore:
= 8,100 / 200
= 40.5 m
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Can someone help me out on these 2 geometry questions pls? ASAP!!!
Write formal proofs using the HL Theorem
Question 2
1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)
2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)
3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)
4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)
5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)
Question 3
1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)
2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)
3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)
4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)
5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)
6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)
7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)
Freddie has 2 times as many comic books as David. The ratio of the number of comic books David has to the number of comic books Gary has is 5 : 3. Freddie has 110 comic books. How many comic books do David and Gary have in total?
Answer:
88
Step-by-step explanation: make me brainliest if it is correct
[tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex]
The solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]
How to solve the expression?The expression is given as:
[tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex]
Take the LCM of the expression
[tex]\frac{8x^3 * 27x^3 - 4x * 27x^3 + 2 * 9x^2 - 1}{27x^3}[/tex]
Evaluate the products
[tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]
Hence, the solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]
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4(3) / 4(8) +.9(7.8 x 7.45)
The value is 52. 674
How to determine the valueWe have;
4(3) / 4(8) +.9(7.8 x 7.45)
From BODMAS, we have that we must must open up the bracket
= [tex]\frac{12}{32} + 0. 9 (58. 11)[/tex]
Divide the fraction to get decimals
= [tex]0. 375 + 52. 299[/tex]
Add the decimals
= [tex]52. 674[/tex]
Thus, the value is 52. 674
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HELPPPPPPPPPPPPP99 n
Answer:
555 meters
Step-by-step explanation:
toymaker = height/shadow
= 0.9/0.2
CN tower = height/shadow
= x/123.3
solve for x:
;[tex]\frac{0.9}{0.2} = \frac{x}{123.3} \\\\0.2(x) = (0.9)(123.3)\\0.2x = 110.97\\(\frac{10}{2} )\frac{2}{10}x = 110.97(\frac{10}{2})\\ x = 554.85[/tex]
They are all in meters, so we don't need to convert anything so rounded to the nearest meter, the answer is 555 meters
Answer: 555 m
Step-by-step explanation:
We will set up a proportion to solve.
[tex]\displaystyle \frac{\text{vertical side}}{\text{horizontal side}} \;\;\;\;\;\;\;\frac{0.9\;m}{0.2\;m} =\frac{x\;m}{123.3 \;m}[/tex]
Now we will cross multiply.
0.9 * 123.3 = 0.2 * x
110.97 = 0.2x
554.85 = x
x = 554.85
Lastly, we will round the nearest m.
554.85 ➜ 555 m
Let [tex]sin\beta =\frac{2\sqrt[]{2} }{5} \\[/tex] and [tex]\frac{\pi }{2} \leq\beta \leq \pi \\[/tex].
Determine the exact value of [tex]sin(\frac{\beta }{2} )[/tex]
Since beta is in the first quadrant, the final answer will be positive.
To find cos(beta) so we can use the half angle identity, we can substitute into the Pythagorean identity. Doing so gives us that
[tex] \sin( \beta ) = \frac{ \sqrt{17} }{5} [/tex]
So, this means that
[tex] \sin( \frac{ \beta }{2} ) = \sqrt{ \frac{1 - \frac{ \sqrt{17} }{2} }{2} } = \sqrt{ \frac{2 - \sqrt{17} }{4} } = \frac{ \sqrt{2 - \sqrt{17} } }{2} [/tex]
Billy goes on the teacup ride at the local carnival. He is seated 16 inches from the center of his teacup, which is spinning at a rate of 4 revolutions per minute. What is Billy’s angular velocity, in radians per second? Round your answer to the nearest hundredth.
Answer:
0.42
Step-by-step explanation:
Five guys walk into a bar, how many ways can they sit so as to be arranged from oldest to youngest?
There are 24 ways in which 5 guys can sit if arranged from oldest to youngest.
We have,
Five guys.
Now,
We know that,
Total number of ways to arranged around a table (n) = (n-1)!
So,
For n = 5,
I.e.
(n - 1)! = (5 - 1)! = 4!
So,
Total number of ways to arranged Five guys rom oldest to youngest (n) = (n-1)!
i.e.
= (5 - 1)! = 4!
We get,
= 4 × 3 × 2 × 1
i.e.
Total number of ways to arranged Five guys rom oldest to youngest (n) = 24.
Hence we can say that there are 24 ways in which 5 guys can sit if arranged from oldest to youngest.
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1. Prove the following identities
Answer:
Step-by-step explanation:
[tex]\sf 1.1)\dfrac{Sin \ \theta-Cos \ \theta}{Sin \ \theta + Cos \ \theta}=\dfrac{(Sin \ \theta-Cos \ \theta)}{((Sin \ \theta + Cos \ \theta)}*\dfrac{(Sin \ \theta-Cos \ \theta)}{(Sin \ \theta-Cos \ \theta)}[/tex]
[tex]\sf = \dfrac{(Sin \ \theta-Cos \ \theta)^2}{Sin^2 \ \theta-Cos^2 \ \theta}\\\\=\dfrac{Sin^2 \ \theta+Cos^2 \ \theta-2Sin \ \theta \ Cos\ \theta}{(Sin \ \theta-Cos \ \theta)}\\\\\\\bf Identity: \ (a +b)^2= a^2 + b^2 - 2ab\\\\\\=\dfrac{1-2Sin \ \theta \ Cos \ \theta}{(Sin \ \theta-Cos \ \theta)} = LHS[/tex]
[tex]\sf 1.2) LHS = tan^2 \ x - Sin^2 \ x = \dfrac{Sin^2 \ x}{Cos^2 \ x}-Sin^2 \ x[/tex]
[tex]\sf =\dfrac{Sin^2 \ x}{Cos^2 \ x}-\dfrac{Sin^2 \ x*Cos^2 \ x}{1*Cos^2 \ x}\\\\\\ = \dfrac{Sin^2 \ x - Sin^2 \ x*Cos^2 \ x}{Cos^2 \ x}\\\\\\= \dfrac{Sin^2 \ x *(1 -Cos^2 \ x)}{Cos^2 \ x}\\\\=\dfrac{Sin^2 \ x*Sin^2 \ x}{Cos^2 \ x} \\\\ \bf 1 - Cos^2 \ x = Sin^2 \ x\\\\= \dfrac{Sin^2 \ x}{Cos^2 \ x}*Sin^2 \ x\\\\=tan^2 \ x * Sin^2 \ x = RHS[/tex]
[tex]\sf 1.3) LHS = \dfrac{1-Cos \ x}{Sin \ x}-\dfrac{Sin \ x}{1+Cos \ x} =\dfrac{(1-Cos \ x)(1+Cos \ x)}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin \ x*Sin \ x}{(1+Cos \ x)*Sin \ x}\\[/tex]
[tex]\sf =\dfrac{1 - Cos^2 \ x}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)} - \dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 x - Sin^2 \ x}{Sin \ x*(1+Cos \ x)} \\\\= 0 = RHS[/tex]
[tex]\sf 1.4) LHS = Sin x - \dfrac{1}{Sin \ x + Cos \ x}+Cos \ x \\[/tex]
[tex]\sf = \dfrac{Sin \ x *(Sin \ x + Cos \ x) - 1 + Cos \ x * (Sin \ x + Cos \ x)}{Sin \ x + Cos \ x }\\\\\\= \dfrac{Sin \ x * Sin \ x + Sin \ x*Cos \ x -1 + Cos \ x*Sin \ x + Cos \ x*Cos \ x}{Sin \ x + Cos \ x}\\\\\\=\dfrac{Sin^2 \x + Sin \ x \ Cos \ x - 1 + Cos \ x \ Sin \ x + Cos^2 \x}{Sin \ x + Cos \ x}\\\\=\dfrac{Sin^2 \ x + Cos^2 \ x - 1 + Sin \ xCos \x +Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{1 - 1 +2Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{2Sin \ x Cos \ x}{Sin \ x + Cos \ x}[/tex]