Answer:
D. She added the wrong value to both sides of the equation to complete the square.
Step-by-step explanation:
Got it right on plato
The first error in Carly's work is instead adding 2² on both the side of an equation, she added 4².
What is the solution of a quadratic equation?The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. The roots of any polynomial are the solutions for the given equation.
The given equation is x(x+4)=117.
Here, x²+4x=117
Now, b/2 =4/2 =2
So, add 2² to both the sides of an equation, we get
x²+4x+2²=117+2²
⇒ x²+4x+4=117+4
⇒ (x+2)²=121
⇒ x+2=±√121
⇒ x+2=±11
⇒ x=±11-2
⇒ x=11-2=9 and x=-11-2=-13
The solution for the given a quadratic equation is x=9 and x=-13. Therefore, the first error in Carly's work is instead adding 2² on both the side of an equation, she added 4².
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can someone pls help me.. thanks <3 Factor 15x^2y^2-3x^3y+75x^4 Show your work.
Answer:
3(x^y(5x^2y-x³)+25x⁴))
Step-by-step explanation:
15x^2y²-3x^3y+75x⁴
From 15x^2y²-3x^3y only, 3x^y is the common factor
=> 3x^y(5x^2y-x³)+75x⁴
Taking the common factor of the latter expression, 3 shows to be the common factor of all the expression.
=> 3(x^y(5x^2y-x³)+25x⁴)
If the length of JK is 3x and the length of LM is 12x and the length of the entire line JM is 25x , find the length of KL.
Answer:
10x.
Step-by-step explanation:
It is given that,
JK = 3x
LM = 12x
JM = 25x
Let as consider the line as shown in the below figure.
From the figure it is clear that,
[tex]JK+KL+LM=JM[/tex]
[tex]3x+KL+12x=25x[/tex]
[tex]KL+15x=25x[/tex]
[tex]KL=25x-15x[/tex]
[tex]KL=10x[/tex]
Therefore, the length of KL is 10x.
A bakery makes 270 scones and 300 muffins each morning. What is the ratio of muffins to scones in simplest form?
Answer: 9 : 10 or 9/10.
Step-by-step explanation:
The ratio of muffins to scones is 300 : 270. Now to reduce it to the lowest term divide each by 30.
You will then get 9: 10
exponents and power - simplify and express the result with positive index
I hope u will get help frm it.....
2. (04.01) Which point could be removed in order to make the relation a function? (4 points) {(-9, -8), (-8, 4), (0, -2), (4, 8), (0, 8), (1, 2)} O (4,8) O (0,8) O (-9, -8) O (1,2)
Answer:
We are given order pairs (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}.
We need to remove in order to make the relation a function.
Step-by-step explanation:
Note: A relation is a function only if there is no any duplicate value of x coordinate for different values of y's of the given relation.
In the given order pairs, we can see that (0, –2) and (0, 8) order pairs has same x-coordinate 0.
So, we need to remove any one (0, –2) or (0, 8) to make the relation a function. hope this helps you :) god loves you :)
Combine the like terms to create an equivalent expression. \large{7n+4n}7n+4n
Answer:
11n
Step-by-step explanation:
The expression is 7n + 4n. Since 7n and 4n are like terms (they both are variables with n), we can combine them so the expression becomes 7n + 4n = 11n.
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has faces numbered 1, 2, 3, 4, 4, and 4. A die is selected at random and rolled four times. You are told that two rolls were 1's and two were 4's. Find the probability the die chosen was green.
Answer:
The probability the die chosen was green is 0.9
Step-by-step explanation:
From the information given :
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has faces numbered 1, 2, 3, 4, 4, and 4.
SO, the probability of obtaining 4 in a single throw of a fair die is:
P (4 | red dice) = [tex]\dfrac{1}{6}[/tex]
P (4 | green dice) = [tex]\dfrac{3}{6}= \dfrac{1}{2}[/tex]
A die is selected at random and rolled four times.
When the die is selected randomly; the probability of the first die must be equal to the probability of the second die = [tex]\dfrac{1}{2}[/tex]
The probability of two 1's and two 4's in the first dice can be calculated as:
[tex]= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4[/tex]
[tex]=\dfrac{4!}{2!(4-2)!}\times (\dfrac{1}{6})^4[/tex]
[tex]=\dfrac{4!}{2!(2)!}\times (\dfrac{1}{6})^4[/tex]
[tex]=\dfrac{4\times 3 \times 2!}{2!(2)!}\times (\dfrac{1}{6})^4[/tex]
[tex]=\dfrac{12}{2 \times 1}\times (\dfrac{1}{6})^4[/tex]
[tex]= 6 \times (\dfrac{1}{6})^4[/tex]
[tex]= (\dfrac{1}{6})^3[/tex]
[tex]= \dfrac{1}{216}[/tex]
The probability of two 1's and two 4's in the second dice can be calculated as:
[tex]= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]
[tex]= \dfrac{4!}{2!(4-2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]
[tex]= \dfrac{4!}{2!(2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]
[tex]= 6 \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]
[tex]= \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]
[tex]= \dfrac{9}{216}[/tex]
∴ The probability of two 1's and two 4's in both dies
= P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )
The probability of two 1's and two 4's = [tex](\dfrac{1}{216} \times \dfrac{1}{2} )+ ( \dfrac{9}{216} \times \dfrac{1}{2})[/tex]
The probability of two 1's and two 4's = [tex]\dfrac{1}{432}+ \dfrac{1}{48}[/tex]
The probability of two 1's and two 4's = [tex]\dfrac{5}{216}[/tex]
Using Bayes Theorem; the probability that the die was green can be computed as follows:
P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's)
P(second die (green) | two 1's and two 4's ) = [tex]\dfrac{\dfrac{1}{2} \times \dfrac{9}{216} }{\dfrac{5}{216}}[/tex]
P(second die (green) | two 1's and two 4's ) = [tex]\dfrac{\dfrac{1}{48} }{\dfrac{5}{216}}[/tex]
P(second die (green) | two 1's and two 4's ) =[tex]\dfrac{1}{48} \times \dfrac{216}{5 }[/tex]
P(second die (green) | two 1's and two 4's ) = [tex]\dfrac{9}{10}[/tex]
P(second die (green) | two 1's and two 4's ) = 0.9
∴
The probability the die chosen was green is 0.9
Let E1 be the event that tails come up when the coin is tossed the first time and E2 be the event that heads come up when the coin is tossed the second time. Drag the probability values from the right column and drop them in the corresponding events in the left column.
Answer:
Step-by-step explanation:
In the tossing of a fair coin, there are equal probabilities of getting a HEAD and getting a TAIL.
Total probability is always 1 and a coin has 2 faces - Head & Tail.
The probability of getting a Head is 1/2 = 0.5
The probability of getting a Tail is 1/2 = 0.5
E1 is the event that TAIL comes up when the coin is tossed the first time
E2 is the event that HEAD comes up when the coin is tossed the second time
The probability value for EVENT 1 is 0.5
The probability value for EVENT 2 is 0.5
Rewrite in simplest terms: -2(5d-9f)+7f-10(-9f-7d)−2(5d−9f)+7f−10(−9f−7d)
Answer:
= 5 ( 12d + 23f )
Step-by-step explanation:
-2(5d-9f)+7f-10(-9f-7d)
Open parenthesis
= -10d + 18f + 7f + 90f + 70d
Collect like terms
= -10d + 70d + 18f + 7f + 90f
= 60d + 115f
Factorise
= 5 ( 12d + 23f )
Therefore,
-2(5d-9f)+7f-10(-9f-7d) in its simplest form is 5 ( 12d + 23f )
pt 3 1-7 pleaseeee helpp
Answer:
-2s is ur answer
hope it helps u
Answer:
Step-by-step explanation:
2s+(−4s)
Combine 2s and −4s to get −2s.
−2s
Find the surface area of the pyramid shown to the nearest whole number.
6 ft
5 ft
5 ft
Not drawn to scale
a. 85 A
b. 145 ft
c. 60 i
d. 25 ft
Answer:
The answer is option AStep-by-step explanation:
Surface area of a pyramid =
area of base + area of triangular faces
Since it's a square based pyramid
It's surface area is
area of base + 4( area of one triangular face)
Since the square has equal sides
For square base
Area of a square = l²
where l is the length
From the question l = 5
So we have
Area of square base = 5² = 25ft²
For one of the triangular face
Area of a triangle = ½ × base × height
base = 5
height = 6
Area = ½ × 5 × 6 = 15ft²
So the surface area of the pyramid is
25 + 4(15)
= 25 + 60
We have the final answer as
Surface area = 85 ft²Hope this helps you
Calculate the rate of change for the table of values.
X
у
2
50
3
75
4.
100
5
125
rate of change =
Answer:
25
Step-by-step explanation:
50:2=25
75:3=25
100:4=25
125:5=25
The rate of change of the given data in the table will be 25.
What is the rate of change?The momentum of a variable is represented by the rate of change, which is used to mathematically express the percentage change in value over a specified period of time.
The formula for the rate of change is straightforward: it simply divides the current value of a stock or index by the value from a previous time period.
The rate of change will be calculated by using the following formula:-
rate of change = ( y₂ - y₁ ) / ( x₂ - x₁ )
The rate of change will be calculated as below:-
50:2=25
75:3=25
100:4=25
125:5=25
Therefore, the rate of change of the given data in the table will be 25.
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what is the volume of the cubic figure? enter the answer in the box below
use the figure to answer the question
9m, 5m, 18m, 10m, 9m, 24m,9m (look at picture)
Answer:
[see below]
Step-by-step explanation:
The figure is made out of two rectangular prisms.
V = w * l * h
Volume for the left prism:
V = 9 * 5 * 24
V = 1080 m³
Volume for the right prism:
V = 9 * 18 * 10
V = 1620 m³
Combine both volumes:
1620 + 1080 = 2700
2700 cubic meters should be your answer.
Hope this helps.
solve by completing the square. 4x²-8x-32=0
Answer:
4, -2
Step-by-step explanation:
Hello, please consider the following.
[tex]\begin{aligned}4x^2-8x-32=0 &\text{ ***We divide by 4.***}\\\\x^2-2x-8=0 &\text{ ***We complete the square. ***}\\\\(x-1)^2-1-8=(x-1)^2-9=0 &\text{ ***We move the constant to the right.***}\\\\(x-1)^2=9=3^2 &\text{ ***We take the root.***}\\\\x-1=\pm3 &\text{ ***We add 1.***}\\\\x=1+3=\boxed{4} \ & or \ x=1-3=\boxed{-2}\end{aligned}[/tex]
Thanks
If AC = 24, find the value of x. Then find AB and BC
Answer:
The answer is below
Step-by-step explanation:
The question is not complete, the correct question is:
If B is between A and C, and AB=3x+1, BC=2x-7, and AC=24, then find the value of x and the value of AB
Answer: The line segment addition postulate states that if a point B is placed between a line segment with end points A and C, then the distance between the points can be expressed by the equation:
AB + BC = AC
But AB=3x+1, BC=2x-7, and AC=24, Hence:
3x + 1 + 2x - 7 = 24
3x + 2x + 1 - 7 = 24
5x - 6 = 24
5x = 24 + 6
5x = 30
x = 6
AB = 3x + 1 = 3(6) + 1 = 18 + 1 = 19
BC = 2x - 7 = 2(6) - 7 = 12 - 7 = 5
Using the cross multiplication method, which of the following is the solution to 21/(x+4) = 7/(x+2)?
Answer:
Step-by-step explanation:
Hello !
Using the cross multiplication method, which of the following is the solution to 21/(x+4) = 7/(x+2)?
21(x + 2) = 7(x + 4)
21x + 42 = 7x + 28
21x - 7x = 28 - 42
14x = -14
x = -14/14
x = -1
Answer:
3/(x+2)= x+1
Step-by-step explanation:
21/7=3
x/x= x
4/2=2
Find f. (Use C for the constant of the first antiderivative, D for the constant of the second antiderivative and E for the constant of the third antiderivative.)
f '''(t) = (t)^1/2 − 9 cos(t)
f(t) = _______.
You just need to integrate 3 times:
[tex]f'''(t)=t^{1/2}-9\cos t[/tex]
[tex]f''(t)=\displaystyle\int f'''(t)\,\mathrm dt=\frac23 t^{3/2}-9\sin t+C[/tex]
[tex]f'(t)=\displaystyle\int f''(t)\,\mathrm dt=\frac4{15} t^{5/2}+9\cos t+Ct+D[/tex]
[tex]f(t)=\displaystyle\int f'(t)\,\mathrm dt=\frac8{105} t^{7/2}+9\sin t+\frac C2 t^2+Dt+E[/tex]
On the first part of her trip Natalie rode her bike 16 miles and on the second part of the trip she rode her bike 42 miles. Her average speed during the second part of the trip was 6 mph faster than her average speed on the first part of the trip. Find her average speed for the second part of the trip if the total time for the trip was 5 hours.
Answer:
14 mph ( average speed during the second part of the trip )
Step-by-step explanation:
Let´s call "x" the average speed during the first part then
t = 5 hours
t = t₁ + t₂ t₁ and t₂ times during part 1 and 2 respectively
l = t*v ( distance is speed by time ) t = l/v
First part
t₁ = 16/x and t₂ = 42 / ( x + 6)
Then
t = 5 = 16/x + 42 /(x + 6)
5 = [ 16 * ( x + 6 ) + 42 * x ] / x* ( x + 6 )
5 *x * ( x + 6 ) = 16*x + 96 + 42 x
5*x² + 30*x - 58*x - 96 = 0
5*x² - 28*x - 96 = 0
We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense
x₁,₂ = [28 ± √ (28)² + 1920 ] / 10
x₁,₂ = ( 28 ± √2704 )/ 10
x₁ = 28 - 52 /10 we dismiss that root
x₂ = 80/10
x₂ = 8 mph average speed during the first part, and the average speed in the second part was 6 more miles than in the firsst part. then the average spedd dring the scond part was 8 + 6 = 14 mph
Simplify the algebraic expression 3+6(−9w+7)
Answer:
[tex] \boxed{ \bold{ \huge{\boxed{ \sf{ - 54w + 45}}}}}[/tex]
Step-by-step explanation:
[tex] \sf{3 + 6( - 9w + 7)}[/tex]
Distribute 6 through the parentheses
⇒[tex] \sf{3 - 54w + 42}[/tex]
Add the numbers : 42 and 3
⇒[tex] \sf{ - 54w + 45}[/tex]
Hope I helped!
Best regards!!
200 is 10 times as much as 20 true
Answer:
yes
Step-by-step explanation:
20x10= 200 therefor its 10times as much
If NO =17 and NP= 5x-6, find they value of x. Then find NP and OP
Greetings from Brasil...
We notice 2 dashes on the NO and NP line. This means that both are the same size. Since NO = 17 then OP is also 17 in length.
So
NP = NO + OP
NP = 17 + 17
NP = 34
As already said
NP = 5X - 6 = 34
5X - 6 = 34
X = 8Arrange the operations into the order they should be performed in.
(60 + 15) - 3-2
Subtract 2.
Add 60 and 15.
Divide by 3.
1: Add 60 and 15.
2: Divide by 3.
3: Subtract 2.
Find the slope of the line that contains the points (4,2) and (7,-4)*
Answer:
-2
Step-by-step explanation:
To find the slope of the line you have to use the equation,
(y2-y1)/(x2-x1)
In this case it is, (-4-2)/7-4)
This simplifies to -2 and this is the slope of the line
Answer:
-8/5
hope this help!
Which of the following equations shows the correct way to apply the Commutative Property of Addition?
Answer:
Commutative Property of Addition: a + b = b + a
Step-by-step explanation:
The Commutative Property of Addition implies that even on changing the order of addition the final result (i.e. the sum) remains the same.
Consider the addition of two numbers, say a and b:a + b = b + a
Suppose a = 5 and b = 6, then:
a + b = 5 + 6 = 11
b + a = 6 + 5 = 11.
Thus, a + b = b + a.
Consider the addition of three numbers, say a, b and c:a + b + c= a + c + b = b + a + c = c + a + b
Suppose a = 4, b = 3 and c = 6, then:
a + b + c = 4 + 3 + 6 = 13
a + c + b = 4 + 6 + 3 = 13
b + a + c = 3 + 4 + 6 = 13
c + a + b = 6 + 4 + 3 = 13.
Thus, a + b + c= a + c + b = b + a + c = c + a + b.
What is the sum of 87 and 234252464375675647
Answer:
234252464375675734
Step-by-step explanation:
addition...
Answer:
here 234252464375734
A survey of over 25,000 Americans aged between 18 and 24 years revealed the following: 88.1% of the 12,678 females and 84.9% of the 12,460 males had high school diplomas.
a) Do the data suggest that females are more likely to graduate from high school than males? Test at a significance level of 5%.
b) Set-up a 95% confidence interval for the difference in the graduation rates between females and males.
c) State the assumptions and conditions necessary for the above inferences to hold.
Answer:
(a) Yes, the data suggest that females are more likely to graduate from high school than males.
(b) A 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .
Step-by-step explanation:
We are given that a survey of over 25,000 Americans aged between 18 and 24 years revealed the following: 88.1% of the 12,678 females and 84.9% of the 12,460 males had high school diplomas.
Let [tex]p_1[/tex] = population proportion of females who had high school diplomas.
[tex]p_2[/tex] = population proportion of males who had high school diplomas.
(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1\leq p_2[/tex] {means that females are less or equally likely to graduate from high school than males}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]p_1 > p_2[/tex] {means that females are more likely to graduate from high school than males}
The test statistics that will be used here is Two-sample z-test statistics for proportions;
T.S. = ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of females having high school diplomas = 88.1%
[tex]\hat p_2[/tex] = sample proportion of males having high school diplomas = 84.9%
[tex]n_1[/tex] = sample of females = 12,678
= sample of males = 12,460
So, the test statistics =
= 7.428
The value of the standardized z-test statistic is 7.428.
Now, at a 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.
Since the value of our test statistics is more than the critical value of z as 7.428 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that females are more likely to graduate from high school than males.
(b) Firstly, the pivotal quantity for finding the confidence interval for the difference in population proportion is given by;
P.Q. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of females having high school diplomas = 88.1%
[tex]\hat p_2[/tex] = sample proportion of males having high school diplomas = 84.9%
[tex]n_1[/tex] = sample of females = 12,678
[tex]n_2[/tex] = sample of males = 12,460
Here for constructing a 95% confidence interval we have used a Two-sample z-test statistics for proportions.
So, 95% confidence interval for the difference in population proportions, ([tex]p_1-p_2[/tex]) is;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < [tex]{(\hat p_1-\hat p_2)-(p_1-p_2)}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ) = 0.95
P( [tex](\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < ([tex]p_1-p_2[/tex]) < [tex](\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ) = 0.95
95% confidence interval for ([tex]p_1-p_2[/tex]) = [[tex](\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] , [tex](\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ]
= [ [tex](0.881-0.849)-1.96 \times {\sqrt{\frac{0.881(1-0.881)}{12,678}+\frac{0.849(1-0.849)}{12,460}} }[/tex] , [tex](0.881-0.849)+1.96 \times {\sqrt{\frac{0.881(1-0.881)}{12,678}+\frac{0.849(1-0.849)}{12,460}} }[/tex] ]
= [0.024, 0.404]
Therefore, a 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .
(c) The assumptions and conditions necessary for the above inferences to hold are;
The data must follow the normal distribution.The sample must be taken from the population data only or the sample represents the population data.Part A What is the area of the blue shaded figure ( π=3.14)? Justify your answer using equations, models, and/or words to explain your mathematical reasoning. Part B What is the perimeter of the blue shaded figure (π=3.14)? Justify your answer using equations, models, and/or words to explain your mathematical reasoning.
Answer:
1372cm²
Step-by-step explanation:
If this were a complete rectangle, it would be 50x40=2000cm². So we need to take that number and subtract the half-circle. A=[tex]\pi[/tex]r² A=3.14*20*20=1256
1256*1/2 = 628
2000-628=1372cm²
Answer:
1372 cm²
Step-by-step explanation:
First, find the area of the rectangle:
A = lw
A = 40(50)
A = 2000 cm²
Next, find the area of the semicircle:
A =([tex]\pi[/tex]r²) / 2
A = (3.14)(20)²
A = 1256/2
A = 628 cm²
Then, subtract the semicircle area from the rectangle's area:
2000 - 628
= 1372 cm²
Represent the following sentence as an algebraic expression, where "a number" is the
letter x.
Twice a number.
Answer:
[tex]x = 2a[/tex]
Step-by-step explanation:
Required
Represent twice a number is x as an algebra
Given that the number is a;
Then
[tex]Twice\ a\ number = 2 * a[/tex]
[tex]Twice\ a\ number = 2a[/tex]
Also,
[tex]Twice\ a\ number = x[/tex]
So, we have that
[tex]x = 2a[/tex]
Hence, the algebraic representation of the given parameters is
[tex]x = 2a[/tex]
A 120$ coat was on sale for 88$. What was the percent of change in the price of this coat.
Answer: about 73 %
Step-by-step explanation:
because 88 of 120 is about 73 %
The percent of change in the price of this coat is 26.66 %.
To find the percent of change in the price of this coat.
What is percentage?A part of a whole expressed in hundredths a high percentage of students attended. Also the result obtained by multiplying a number by a percent the percentage equals the rate times the base.
Given that:
Cost price of coat(C.P)= 120$
Selling price of coat(S.P)=88$
We know that ,
Percent change=change in price /C.P*100
Percent change=120$-88$/120$*100
Percent change=32/120*100
Percent change=26.66 %
Therefore, the percent of change in the price of this coat is 26.66 %.
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who was the second president in Republican of the congo?
please I need you answer
Answer: Joseph Kabila
Step-by-step explanation:
Answer:
Joseph Kabila
Step-by-step explanation: