Answer:
B
Introduction:Given equation: [tex]x^2 = -8x -7[/tex]
To solve the quadratic equation, we will need to set one side of the equation equal to 0. This can be done by subtracting x² on both sides.
[tex]\implies x^2 - x^2= -8x -7 - x^2[/tex]
[tex]\implies 0= -8x -7 - x^2[/tex]
To further simplify, we can factor -1 from the expression on the R.H.S. Then, we can cancel the -1 since there is a 0 on the other side.
[tex]\implies 0= -1(8x + 7 + x^2)[/tex]
[tex]\implies 0= 8x + 7 + x^2 = x^2 + 8x + 7[/tex]
Finally, factor the expression on the right hand side.
Factoring a quadratic equation can be done in 2 ways.
1) Splitting methodIn this method, we split the middle term into two suitable terms. The suitable terms must follow two conditions.
The product of the suitable terms must be equal to the product of the first and last terms of the expression. The sum of the suitable terms must be equal the middle term.Then, we factor the first two terms of the expression and the last two terms of the expression. Eventually, we will get a common expression. So, we can further factorize to get a product of two expressions.
However, this method follows a condition. If the middle term cannot be split into two suitable terms, then the second method should be followed to determine a solution. 2) The Quadratic FormulaIn this method, we will compare our equation with the standard form equation to determine the a, b, and c variables. Then, we plug the variables into the formula and simplify to get the roots (solutions).
[tex]\text{Quadratic Formula:} \ \dfrac{-b\ \± \sqrt{b^{2} - 4ac} }{2a}[/tex]
We can use any method to solve for the roots. First, let us check whether the middle term could be split into two suitable terms such that those terms could meet the criteria stated above.
-----------------------------------------------------------------------------------------------------
Solving for x using the splitting methodWe can see that 8x is the middle term and it can be split into the sum of 7x and x. The product of the (middle term) coefficients is 7. The product of the coefficients (first and last terms) is also 7. Since both products match, we can use the splitting method to factorize the expression.
[tex]\implies 0= x^2 + 8x+ 7[/tex]
[tex]\implies 0= x^2 + x + 7x+ 7[/tex]
Now, factor the first two terms and the last two terms separately.
[tex]\implies 0= x^2 + x + 7x+ 7[/tex]
[tex]\implies 0= x(x + 1) + 7(x+ 1)[/tex]
We can see a common expression. Finally, let us factor the common expression to get a product of two expressions.
[tex]\implies 0= (x + 7)(x + 1)[/tex]
Since the factorized expression equals to 0, we can set each expression (inside the parentheses) equal to 0.
[tex]\implies x + 7 = 0; x + 1 = 0[/tex]
[tex]\implies x = -7; x = -1[/tex]
Solving for x using the Quadratic formulaNow, let us try to solve using the Quadratic formula.
[tex]0= x^2 + 8x+ 7[/tex]
Tip: Standard form: 0 = ax² + bx + c
Comparing both equations will allow us to determine the a, b, c variables.
⇒ a = 1; b = 8; c = 7
Now, let us plug these values into the quadratic formula and simplify.
[tex]\implies \dfrac{-b\ \± \sqrt{b^{2} - 4ac} }{2a}[/tex]
[tex]\implies \dfrac{-8\ \± \sqrt{8^{2} - 4(1)(7)} }{2(1)}[/tex]
[tex]\implies \dfrac{-8\ \± \sqrt{64 - 28} }{2}[/tex]
[tex]\implies \dfrac{-8\ \± \sqrt{36}}{2} = \dfrac{-8\± 6}{2} = \dfrac{-8 + 6}{2},\ \dfrac{-8 - 6}{2} = \dfrac{-2}{2} ,\ \dfrac{-14}{2} = -1 , -7 \ \text{(B)}[/tex]
Therefore, in both methods, the answer is -1, -7 (B).
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Three circles are inscribed in a rectangle of width w and height h as shown. Two of the circles are congruent and are each tangent to two adjacent sides of the rectangle and to each other. The other circle is larger and is tangent to three sides of the rectangle and to the two smaller circles. What the ratio of h to w? Express your answer as a decimal to the nearest hundredth.
==========================================
Explanation:
The two smaller circles have a height of h, so one circle has a height of h/2 = 0.5h
The radius of each smaller circle is (0.5h)/2 = 0.25h
Draw an xy axis. Place the bottom left corner of the rectangle at the origin (0,0)
The center of the lower smaller circle is at location (0.25h, 0.25h). Call this point A.
Let B be the center of the larger circle. It has coordinates (x,y). We don't know x, but we know that y = 0.5h since the center must be at the halfway point in terms of the height of this rectangle. So the larger circle has a radius of 0.5h
Draw a line segment connecting A and B. The length of this segment, call it d, is d = 0.5h + 0.25h = 0.75h. Note how I added the two radius values mentioned earlier.
-------------
Summarizing everything so far, we have
A = (0.25h, 0.25h)
B = (x, 0.5h)
d = 0.75h
The distance formula is then used
d = distance from A to B
d = length of segment AB
d = sqrt( (x1-x2)^2 + (y1-y2)^2 )
0.75h = sqrt( (0.25h - x)^2 + (0.25h - 0.5h)^2 )
(0.75h)^2 = (0.25h - x)^2 + (-0.25h)^2
0.5625h^2 = 0.0625h^2 - 0.5hx + x^2 + 0.0625h^2
x^2 - 0.5hx + 0.4375h^2 = 0
From here you use the quadratic formula to get x = 0.9571067811865h approximately (the other solution is ignored as it's negative). See the attached image below if you're curious what the quadratic formula steps would look like.
This x value is the x coordinate of point B, which is the center of the larger circle. This spans the horizontal distance from the left edge of the rectangle to the center of the larger circle. The remaining horizontal distance is h/2 as it is the radius of the larger circle.
Therefore,
w = 0.9571067811865h + 0.5h
w = 1.4571067811865h
-------------
We have turned w into a roughly equivalent expression that has an h in it, allowing us to find the ratio of h to w
h/w = h/(1.4571067811865h) = 1/1.4571067811865 = 0.68629150101527
When rounding to two decimal places, we get roughly 0.69
What are the roots of the equation x ^2+ 6x + 8 = 0?
Hi there! :)
Answer:
[tex]\huge\boxed{x = -4, -2}[/tex]
x² + 6x + 8 = 0
Factor the equation by finding two numbers that sum up to 6 and multiply into 8:
4 + 2 = 6
4 ×2 = 8
Rewrite in factored form:
(x + 4) (x + 2) = 0
Find the roots using the Zero-Product Property:
x + 4 = 0
x = - 4
x + 2 = 0
x = -2.
Therefore, the roots of this equation are at x = -4 and -2.
solve (x-5)^2=3 question has to be 23 characters so I'm typing this lol
hi my litler friend
[tex](x-5)^2=3\\(x-5)=+-\sqrt{3} \\\\x-5=\sqrt{3} \\x=5+\sqrt{3} \\\\x-5=-\sqrt{3} \\x=5-\sqrt{3} \\\\\\S=(5-\sqrt{3},5+\sqrt{3})\\[/tex]
if in doubt, let me know. I really like to help.
Exact solutions: [tex]x = 5+\sqrt{3} \ \text{ or } \ x = 5-\sqrt{3}[/tex]
Approximate solutions: [tex]x \approx 6.732051 \ \text{ or } \ x \approx 3.267949[/tex]
==================================================
Work Shown:
Apply the square root to both sides, then add 5 to both sides
Don't forget about the plus/minus so you can get two solutions.
[tex](x-5)^2 = 3\\\\\sqrt{(x-5)^2} = \sqrt{3}\\\\|x-5| = \sqrt{3}\\\\x-5 = \pm\sqrt{3}\\\\x-5+5 = \pm\sqrt{3}+5\\\\x = 5\pm\sqrt{3}\\\\x = 5+\sqrt{3} \ \text{ or } \ x = 5-\sqrt{3}\\\\x \approx 6.732051 \ \text{ or } \ x \approx 3.267949\\\\[/tex]
Use a calculator for the last step.
What is the answer to (-18) - 14 =
Answer:
-32
Step-by-step explanation:
Negative plus negative = Negative
Just add those 2 together,
(-)18 + (-) 14
Which is -32
(-) 32 = -32
Drag each expression to the correct location on the solution. Not all expressions will be used.
Answer:
Hey there!
[tex]8x^3+2x^2-20x-5[/tex]
[tex](8x^3+2x^2)+(-20x-5)[/tex]
[tex]2x^2(4x+1)+(-5)(4x+1)[/tex]
[tex](2x^2-5)(4x+1)[/tex]
Let me know if this helps :)
The solution of the expression are,
⇒ (4x + 1) (2x² - 5)
What is an expression?Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.
Given that;
The expression is,
⇒ 8x³ + 2x² - 20x - 5
Now, We can simplify as;
⇒ 8x³ + 2x² - 20x - 5
⇒ (8x³ + 2x² ) + (- 20x - 5)
⇒ 2x² (4x + 1) - 5 (4x + 1)
⇒ (4x + 1) (2x² - 5)
Thus, The solution of the expression are,
⇒ (4x + 1) (2x² - 5)
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convert 12500 euros into cedis at rate of 1 euro is equal to 4.9 cedis
Answer:
12,500 euro = 61,250 cedis
Step-by-step explanation:
1 euro = 4.9 cedis
12500 euro = 4.9 * 12500 cedis
12,500 euro = 61,250 cedis
Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20,$ how many possible values of $n$ are there?
Answer:
18 values for n are possible.
Step-by-step explanation:
Given the quadratic polynomial:
[tex]$x^2 - mx + n$[/tex]
such that:
Roots are positive prime integers and
[tex]$m < 20$[/tex]
To find:
How many possible values of [tex]n[/tex] are there ?
Solution:
First of all, let us have a look at the sum and product of a quadratic equation.
If the quadratic equation is:
[tex]Ax^{2} +Bx+C[/tex]
and the roots are: [tex]\alpha[/tex] and [tex]\beta[/tex]
Then sum of roots, [tex]\alpha+\beta = -\frac{B}{A}[/tex]
Product of roots, [tex]\alpha \beta = \frac{C}{A}[/tex]
Comparing the given equation with standard equation, we get:
A = 1, B = -m and C = n
Sum of roots, [tex]\alpha+\beta = -\frac{-m}{1} = m[/tex]
Product of roots, [tex]\alpha \beta = \frac{n}{1} = n[/tex]
We are given that [tex]m<20[/tex]
[tex]\alpha[/tex] and [tex]\beta[/tex] are positive prime integers such that their sum is less than 20.
Let us have a look at some of the positive prime integers:
2, 3, 5, 7, 11, 13, 17, 23, 29, .....
Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.
So, possible combinations and possible value of [tex]n (= \alpha \times \beta)[/tex] are:
[tex]1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\[/tex]
[tex]11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77[/tex]
So,as shown above 18 values for n are possible.
5 Seth bought a used car that had been driven 20,000
miles. After he owned the car for 2 years, the total
mileage of the car was 49,400. Find the average
number of miles he drove each month during those
2 years.
49,400-20,000=29400
29400/24=1225 average miles per month
please help me with this problem :))
Answer:
ONE IS DA ANSWR 1
Step-by-step explanation:
so 5/12 +5/6 simpify 5/6 you get 5/12 + 10/12 = 15/12 and 15/12 is 1 and 3/12 1/3
Which of the following interpretations for the given expression is correct?
Answer:
Step-by-step explanation:
Hello friend!!!!!
Third option is the correct answer.
Hope this helps
plz mark as brainliest!!!!!!!
Step-by-step explanation:
Hey, there!!
Option 3 is the correct answer.
here,
when we cube the sum of 5 and 3 times y^2 it looks like,
=[tex] {(5 + 3 {y}^{2}) }^{3} [/tex]
now, when we divide it by cube of the product of 4 and y we get,
=[tex] \frac{ {(5 + 3 {y}^{2}) }^{3} }{ {4y}^{3} } [/tex]
so, answer is option 3.
Hope it helps..
please help me on this geometry!!!
Answer:
∠ PQR = 143°
Step-by-step explanation:
∠ PQR and ∠ SQR are adjacent angles and are supplementary, thus
∠ PQR = 180° - 37° = 143°
use the graph to determine the domain and range of the relation and state whether the relation is a function
Answer:
the domain is {0, 1, 2, 3, 4... to positive infinity}
The range is {-1, -3, 0, -4 and so on}
this is not a function
Step-by-step explanation:
domain is a set of all x values, so just write the x values based on the graph
range is a set of all y values, so just write y values based on the graph
not function because it doesnt satisfy with the vertical line test.
i hop it help u
if m<FHI=142, what are m<FHG and m<GHI
angle FHG = 42 degrees
angle GHI = 100 degrees
=========================================================
Explanation:
FHI is the largest angle. It is split into two pieces FHG and GHI which have measures of (3x+6) and (9x-8) degrees respectively.
Put another way, those two smaller angles (FHG and GHI) combine to form the larger angle (FHI)
This is the angle addition postulate.
------------------
(angle FHG) + (angle GHI) = angle FHI
(3x+6) + (9x-8) = 142
3x+6+9x-8 = 142
(3x+9x) + (6-8) = 142
12x - 2 = 142
12x-2+2 = 142+2 ... adding 2 to both sides
12x = 144
12x/12 = 144/12 .... dividing both sides by 12
x = 12
------------------
Use that x value to find the angles we're after
angle FHG = 3x+6 = 3*12+6 = 36+6 = 42
angle GHI = 9x-8 = 9*12-8 = 108-8 = 100
Note how
(angle FHG) + (angle GHI) = 42 + 100 = 142
which is the measure of angle FHI. This confirms the answers.
Theresa runs 2/7 of a mile in 1/4 of an hour. How fast does she run per hour?
Answer:
15/28
Step-by-step explanation:
How do you tell when the slope is negative?
If the line goes downhill as you read from left to right, then the slope is negative.
The slope is positive when you go uphill, reading from left to right.
A slope of 0 means the line is completely horizontal or level
If the line is completely vertical, then the slope is undefined.
big boy points for big boy question
Answer:
36[tex]\sqrt{3}[/tex] + 72
Step-by-step explanation:
The length of the rectangle = diameter of 2 circles = 4 × 3 = 12 mm
The width of the rectangle = 2 radii + distance d between radii ( lower to upper )
Using the right triangle formed by joining the 3 centres with legs 3 and d and hypotenuse 6
Using Pythagoras' identity, then
d² + 3² = 6²
d² + 9 = 36 ( subtract 9 from both sides )
d² = 27 ( take the square root of both sides )
d = [tex]\sqrt{27}[/tex] = 3[tex]\sqrt{3}[/tex]
Thus width of rectangle = 3 + 3[tex]\sqrt{3}[/tex] + 3 = 6 + 3[tex]\sqrt{3}[/tex]
Hence
area = length × width
= 12 × (6 + 3[tex]\sqrt{3}[/tex] ) = 36[tex]\sqrt{3}[/tex] + 72 mm ²
36 [tex]\sqrt{3}[/tex]+ 72
The length of the rectangle = diameter of 2 circles = 4 × 3 = 12 mm
The width of the rectangle = 2 radii + distance d between radii ( lower to upper )
Using the right triangle formed by joining the 3 centres with legs 3 and d and hypotenuse 6
Using Pythagoras' identity, then
d² + 3² = 6²
d² + 9 = 36 ( subtract 9 from both sides )
d² = 27 ( take the square root of both sides )
d = = 3
Thus width of rectangle = 3 + 3 + 3 = 6 + 3
Hence
area = length × width
= 12 × (6 + 3 ) = 36 + 72 mm ²
If DF = 9x - 39, find EF
Answer:
EF = 58 units
Step-by-step explanation:
Given:
DF = 9x - 39
DE = 47
EF = 3x + 10
Find:
EF
Computation:
DF = DE + EF
9x - 39 = 47 + 3x + 10
9x - 3x = 47 + 39 + 10
6x = 96
x = 16
EF = 3x + 10
EF = 3(16) + 10
EF = 58 units
joe is responsible for reserving hotel rooms for a company trip. His company changes plans and increases how many people are going on the trip, so they need at least 50 5050 total rooms. Joe had already reserved and paid for 16 1616 rooms, so he needs to reserve additional rooms. He can only reserve rooms in blocks, and each block contains 8 88 rooms and costs $ 900 $900dollar sign, 900. Let B BB represent the number of additional blocks that Joe reserves. 1) Which inequality describes this scenario? Choose 1 answer: Choose 1 answer: (Choice A) A 16 + 8 B ≤ 50 16+8B≤5016, plus, 8, B, is less than or equal to, 50 (Choice B) B 16 + 8 B ≥ 50 16+8B≥5016, plus, 8, B, is greater than or equal to, 50 (Choice C) C 16 + B ≤ 50 16+B≤5016, plus, B, is less than or equal to, 50 (Choice D) D 16 + B ≥ 50 16+B≥5016, plus, B, is greater than or equal to, 50 2) What is the least amount of additional money Joe can spend to get the rooms they need? dollars
The inequality expression is :
16 + 8B ≥ 50
Least amount that could be spent = $4500
The total number of rooms required should be atleast 50
Already booked = 16
Number of rooms per block = 8
If the number of blocks = B
Summing together :
Then we have :
16 + 8B ≥ 50
The leat additional amount to be spent:
(50 - 16) / 8
34/8 = 4.25
Least number of blocks, B = 5
Least amount = $900 * 5 = $4500
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Inequalities are used to represent unequal expressions.
The inequality that represents the scenario is: [tex]8B + 16 \ge 50[/tex].The least additional amount is $4500Given that:
[tex]Room = 50[/tex] at least
[tex]Reserved = 16[/tex]
[tex]Rate = 8[/tex] per block
A. The inequality
Let the number of blocks be B.
"At least" means, greater than or equal to. So, we have:
[tex]Rate \times B + Reserved \ge 50[/tex]
Substitute known values
[tex]8 \times B + 16 \ge 50[/tex]
[tex]8B + 16 \ge 50[/tex]
So, the inequality that describes the scenario is:
[tex](b).\ 8B + 16 \ge 50[/tex]
B. The least amount
Solve for B in: [tex]8B + 16 \ge 50[/tex]
[tex]8B \ge 50 - 16[/tex]
[tex]8B \ge 34[/tex]
Divide both sides by 8
[tex]B \ge 4.25[/tex]
The number of blocks cannot be a decimal. So, we round 4.25 to the smallest integer greater than 4.25
This gives:
[tex]B \ge 5[/tex]
The least number of block is 5.
Given that:
[tex]Cost = \$900[/tex] per block
The least amount is calculated as follows:
[tex]Least = B \times Cost[/tex]
So, we have:
[tex]Least = 5 \times \$900[/tex]
[tex]Least = \$4500[/tex]
Hence, the least additional money is: $4500
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Three sides of a triangle are 2 + 2 + 2, 4 + 52 − 32 + 2. Find
its perimeter
Answer:
2+2+2=6
4+52-32+2=26
finally add 26+6=32
What is the point of comparing prices?
Answer:
To know which is better to pay.
Step-by-step explanation:
Good Luck
What is the value of x in the equation below?
R
-3-(-8)-(-2)= X
0-13
-9
03
O 7
Answer:
X= 7
Step-by-step explanation:
In a class of 40 students, 30 read chemistry and 20 read physics. If all the students read at least one if the subjects, find the probability that a student selected at rand from the class reads only chemistry
hi
40 students
30 read chemistry
20 read physics
so logical deduction is that 10 read chemistry and physics.
So only 20 read only chemistry and only 10 read only physics.
So read only chemistry is 20/40 = 1/2
Answer as a fraction = 1/2
Answer in decimal form = 0.5
Answer in percent form = 50%
All three represent the same idea, but in a different form.
=======================================================
Work Shown:
30 read chem, 20 read physics
30+20 = 50 read either one, but this is over 40, so 50-40 = 10 students must read both
From that,
30-10 = 20 read chem only
20-10 = 10 read physics only
This makes the venn diagram you see below
Divide the number of people who read chemistry only (20) over the total number of people (40). This gives 20/40 = 2/4 = 1/2
Then you would use your calculator to find 1/2 = 0.5 = 50%
I NEED HELP ASAP! BRAINLIEST IS BEING GIVEN OUT! What are the solutions of |3x + 2| > 9? A. x > –113 or x > 73 B. x 73 C. x > –113 or x < 73 D. x < –113 or x < 73
Answer:
x > 7/3 or x < -11/3
Step-by-step explanation:
|3x + 2| > 9
There are two solutions, one positive and one negative, remembering to flip the inequality when taking the negative solutions
3x+2 > 9 or 3x+2 < -9
Subtract 2
3x+2-2 >9-2 or 3x+2-2 < -9-2
3x >7 3x>-11
Divide by 3
3x/3 > 7/3 3x/3 < -11/3
x > 7/3 or x < -11/3
- X= 7
The equation has one solution
A value of x that makes the equation true is______
which when simplified makes
the equation turn into___=___
A value of x that makes the equation false is___
which when simplified makes
the equation turn into___=___
Answer:
To make a true equation, check your math to make sure that the values on each side of the equals sign are the same. Ensure that the numerical values on both sides of the "=" sign are the same to make a true equation. For example, 9 = 9 is a true equation. 5 + 4 = 9 is a true equation.You will be able to determine if an equation has one solution (which is when one variable equals one number), or if it has no solution (the two sides of the equation are not equal to each other) or infinite solutions (the two sides of the equation are identical).
GEOMETRY
SOMEBODY PLEASE HELP ME ON THIS I NEED IT BY TODAY
What is the correct postulate/theorem/definition
given: angle MRS = angle MRO
VO=VO
This is the idea that any segment is the same length as itself (in this case, segment VO is the same length as segment VO). Think of a mirror reflecting an identical image of something in front of the mirror.
It might seem redundant to say VO = VO, but it's useful when you want to break up a figure into smaller more manageable pieces to do the proof.
The measure of angle MRS exists equivalent to the measure of the angle MRO or m∠MRS = m∠MRO.
What is quadrilateral?It exists described as the four-sided polygon in geometry containing four edges and four corners.
∠MRS ≅ ∠MRO (given)
From the given figure, we get
∠MRS + ∠MRO = ∠SRO
∠MRS + ∠MRO = 180 (as SRO exists a straight line segment)
Given: ∠MRS exists congruent to ∠MRO
2∠MRS = 180 degree
∠MRS = 90 degree
∠MRO = 180 - 90 = 90 degree
m∠MRS = m∠MRO
Therefore, the measure of angle MRS exists equivalent to the measure of the angle MRO or m∠MRS = m∠MRO.
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Esteban finds 12 baseball cards on Monday and 24 more baseball cards on Tuesday. He gives some baseball cards to his sister, and now he has 13 baseball cards left. How many cards did he give to his sister?
Answer: 23
Step-by-step explanation: he had 12 and 24 which makes 36
then he had 13 which means he gave 23 because adding them together gives the original amount
What is the degree of
5x^2 -4x+2x^3
Answer:
This is a third-degree trinomial.
Step-by-step explanation:
To find the degree of a polynomial, simply look at the term with the highest power. In the case of this polynomial, the term with the highest power is 2x^3 which is a third power. Hence, this is a 3rd degree trinomial.
Cheers.
184 sixth graders are going on a field trip. There needs to be one chaperone for every four students. If a bus can hold 50 people, how many busses will they need for the trip? Hint: You can't use part of a bus! *
Answer:
5 buses
Step-by-step explanation:
Total number of 6th graders going for the field trip = 184
Step 1
Divide the total number of students by 4
184/4 = 46
We have 46 groups of six graders.
Step 2
There needs to be one chaperone for every four students.
We have 46 groups of six graders
This means there is need for 1 chaperone per group
Hence, we have 46 chaperones
Step 3
Find total number of people going for the field trip
184 students + 46 chaperones
= 230 people
Step 4
If a bus can hold 50 people, how many buses will they need for the trip?
50 people = 1 bus
230 people = ??y
50 × y = 230
y = 230/50
y = 4.6 buses
Hence, 4.6 buses would be needed but since we can't have 0.6(part) of a bus figuratively, as buses come in whole numbers, we approximate to the nearest whole number
= 4.6 ≈ 5
Therefore, they would be needing 5 buses for the trip.
5 buses
Step-by-step explanation:
i got 10 out of 10
X - 25 is a factor of 4x^2 -100?
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Hi my lil bunny!
❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙
Answer : [tex]\boxed{(2x + 5 ) (2x - 5)}[/tex]
[tex]4x^2-25[/tex]
Rewrite [tex]4x^2[/tex] as [tex](2^x)^2[/tex].
[tex](2x)^2 - 5^2[/tex]
Rewrite [tex](2x)^2[/tex] as [tex]5^2[/tex].
[tex](2x)^2 - 5^2[/tex]
Since both terms are perfect squares, factor using the difference of squares formula, [tex]a^2 - b^2 = ( a + b ) ( a - b )[/tex] where [tex]a = 2x[/tex] and [tex]b = 5[/tex].
[tex](2x + 5 ) (2x - 5)[/tex]
❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙
●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●
Have a great day/night!
❀*May*❀
23891-18939 answer pls thanks
Merhaba ;))
Answer:
4.952
#Başarılar
Answer:
4952Step-by-step explanation:
[tex]\begin{matrix}\space\space&2&3&8&9&\textbf{1}\\ -&1&8&9&3&\textbf{9}\end{matrix}\\\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:\\Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}\\\mathrm{Borrow\:}1\mathrm{\:from\:}9\mathrm{.\:\:The\:remainder\:is\:}8\\\begin{matrix}\space\space&\space\space&\space\space&\space\space&\textbf{8}&10\\ \space\space&2&3&8&\textbf{\linethrough{9}}&1\\ -&1&8&9&\textbf{3}&9\end{matrix}[/tex]
[tex]\begin{matrix}\space\space&\space\space&\space\space&\space\space&8&\textbf{11}\\ \space\space&2&3&8&\linethrough{9}&\textbf{\linethrough{1}}\\ -&1&8&9&3&\textbf{9}\end{matrix}[/tex]
[tex]\frac{\begin{matrix}\space\space&\space\space&\space\space&\space\space&\textbf{8}&11\\ \space\space&2&3&8&\textbf{\linethrough{9}}&\linethrough{1}\\ -&1&8&9&\textbf{3}&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\space\space&\textbf{5}&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&\space\space&\textbf{\space\space}&\space\space&8&11\\ \space\space&2&\textbf{3}&8&\linethrough{9}&\linethrough{1}\\ -&1&\textbf{8}&9&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{\space\space}&\space\space&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&\textbf{1}&10&\space\space&8&11\\ \space\space&\textbf{\linethrough{2}}&3&8&\linethrough{9}&\linethrough{1}\\ -&\textbf{1}&8&9&3&9\end{matrix}}{\begin{matrix}\space\space&\textbf{\space\space}&\space\space&\space\space&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&1&\textbf{13}&\space\space&8&11\\ \space\space&\linethrough{2}&\textbf{\linethrough{3}}&8&\linethrough{9}&\linethrough{1}\\ -&1&\textbf{8}&9&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{\space\space}&\space\space&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&1&\textbf{12}&10&8&11\\ \space\space&\linethrough{2}&\textbf{\linethrough{13}}&8&\linethrough{9}&\linethrough{1}\\ -&1&\textbf{8}&9&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{\space\space}&\space\space&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&1&12&\textbf{18}&8&11\\ \space\space&\linethrough{2}&\linethrough{13}&\textbf{\linethrough{8}}&\linethrough{9}&\linethrough{1}\\ -&1&8&\textbf{9}&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{\space\space}&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&1&12&\textbf{18}&8&11\\ \space\space&\linethrough{2}&\linethrough{13}&\textbf{\linethrough{8}}&\linethrough{9}&\linethrough{1}\\ -&1&8&\textbf{9}&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{9}&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&1&\textbf{12}&18&8&11\\ \space\space&\linethrough{2}&\textbf{\linethrough{13}}&\linethrough{8}&\linethrough{9}&\linethrough{1}\\ -&1&\textbf{8}&9&3&9\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{4}&9&5&2\end{matrix}}[/tex]
[tex]\frac{\begin{matrix}\space\space&\textbf{1}&12&18&8&11\\ \space\space&\textbf{\linethrough{2}}&\linethrough{13}&\linethrough{8}&\linethrough{9}&\linethrough{1}\\ -&\textbf{1}&8&9&3&9\end{matrix}}{\begin{matrix}\space\space&\textbf{0}&4&9&5&2\end{matrix}}\\\\=4952[/tex]