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3 years ago

7

Kendra was asked to find the number of students who planned to attend the evening sympothy performance. She was told that 40 per

cent of the 50 students planned to attend. Kendra's work is shown below.
Part 40x2 80
------- = --------- = -------
whole 100 x 2 200

What mistake did Kendra make?
A. Kendra should have multiplied by 4 instead of multiplying by 2.
B. Kendra should have divided by 2 instead of multiplying by 2.
C. Kendra should have added 40 and 50.
D. Kendra should have subtracted 40 from 50.

ILL GIVE BRAINLIEST TO THE RIGHT ANSWER!!!!!!

2 answers:

Paha777 [63]3 years ago

5 0

Answer:

Hey Deku it's me AnimeLover10

Step-by-step explanation:

ch4aika [34]3 years ago

4 0

Answer:

It's B

Step-by-step explanation:

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Will reward brainliest What is the value of x?

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Answer:

x=5

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If 27^27+27^27+27^27=3^k<br> What is k?

JulijaS [17]

Answer:

$\large{\boxed{k=82}}$

Step-by-step explanation:

$27^{27}+27^{27}+27^{27}=3^k$

$3 \cdot 27^{27}=3^k$

$3 \cdot (3^3)^{27}=3^k$

$3 \cdot 3^{81}=3^k$

$3^{82}=3^k \implies k=82$

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3 years ago

Find the area of the 4 walls of room whose length is 15, breadth is 12 and the height is 14.

Whitepunk [10]

Answer:756

Step-by-step explanation:

the formula of the four walls is 2(lh+bh)

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3 years ago

If it exists, solve for the inverse function of each of the following:

nata0808 [166]

Answer:

<em>The solution is too long. So, I included them in the explanation</em>

Step-by-step explanation:

This question has missing details. However, I've corrected each question before solving them

Required: Determine the inverse

1:

$f(x) = 25x - 18$

Replace f(x) with y

$y = 25x - 18$

Swap y & x

$x = 25y - 18$

$x + 18 = 25y - 18 + 18$

$x + 18 = 25y$

Divide through by 25

$\frac{x + 18}{25} = y$

$y = \frac{x + 18}{25}$

Replace y with f'(x)

$f'(x) = \frac{x + 18}{25}$

2. $g(x) = \frac{12x - 1}{7}$

Replace g(x) with y

$y = \frac{12x - 1}{7}$

Swap y & x

$x = \frac{12y - 1}{7}$

$7x = 12y - 1$

Add 1 to both sides

$7x +1 = 12y - 1 + 1$

$7x +1 = 12y$

Make y the subject

$y = \frac{7x + 1}{12}$

$g'(x) = \frac{7x + 1}{12}$

3: $h(x) = -\frac{9x}{4} - \frac{1}{3}$

Replace h(x) with y

$y = -\frac{9x}{4} - \frac{1}{3}$

Swap y & x

$x = -\frac{9y}{4} - \frac{1}{3}$

Add $\frac{1}{3}$ to both sides

$x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}$

$x + \frac{1}{3}= -\frac{9y}{4}$

Multiply through by -4

$-4(x + \frac{1}{3})= -4(-\frac{9y}{4})$

$-4x - \frac{4}{3}= 9y$

Divide through by 9

$(-4x - \frac{4}{3})/9= y$

$-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y$

$\frac{-4x}{9} - \frac{4}{27}= y$

$y = \frac{-4x}{9} - \frac{4}{27}$

$h'(x) = \frac{-4x}{9} - \frac{4}{27}$

4:

$f(x) = x^9$

Replace f(x) with y

$y = x^9$

Swap y with x

$x = y^9$

Take 9th root

$x^{\frac{1}{9}} = y$

$y = x^{\frac{1}{9}}$

Replace y with f'(x)

$f'(x) = x^{\frac{1}{9}}$

5:

$f(a) = a^3 + 8$

Replace f(a) with y

$y = a^3 + 8$

Swap a with y

$a = y^3 + 8$

Subtract 8

$a - 8 = y^3 + 8 - 8$

$a - 8 = y^3$

Take cube root

$\sqrt[3]{a-8} = y$

$y = \sqrt[3]{a-8}$

Replace y with f'(a)

$f'(a) = \sqrt[3]{a-8}$

6:

$g(a) = a^2 + 8a- 7$

Replace g(a) with y

$y = a^2 + 8a - 7$

Swap positions of y and a

$a = y^2 + 8y - 7$

$y^2 + 8y - 7 - a = 0$

Solve using quadratic formula:

$y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

$a = 1$ ; $b = 8$ ; $c = -7 - a$

$y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$ becomes

$y = \frac{-8 \±\sqrt{8^2 - 4 * 1 * (-7-a)}}{2 * 1}$

$y = \frac{-8 \±\sqrt{64 + 28 + 4a}}{2 * 1}$

$y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}$

$y = \frac{-8 \±\sqrt{92 + 4a}}{2 }$

Factorize

$y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }$

$y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }$

$y = -4 \±\sqrt{(23 + a)}$

$g'(a) = -4 \±\sqrt{(23 + a)}$

7:

$f(b) = (b + 6)(b - 2)$

Replace f(b) with y

$y = (b + 6)(b - 2)$

Swap y and b

$b = (y + 6)(y - 2)$

Open Brackets

$b = y^2 + 6y - 2y - 12$

$b = y^2 + 4y - 12$

$y^2 + 4y - 12 - b = 0$

Solve using quadratic formula:

$y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

$a = 1$ ; $b = 4$ ; $c = -12 - b$

$y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$ becomes

$y = \frac{-4\±\sqrt{4^2 - 4 * 1 * (-12-b)}}{2*1}$

$y = \frac{-4\±\sqrt{4^2 - 4 *(-12-b)}}{2}$

Factorize:

$y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}$

$y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}$

$y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}$

$y = \frac{-4\±2\sqrt{16+b}}{2}$

$y = -2\±\sqrt{16+b}$

Replace y with f'(b)

$f'(b) = -2\±\sqrt{16+b}$

8:

$h(x) = \frac{2x+17}{3x+1}$

Replace h(x) with y

$y = \frac{2x+17}{3x+1}$

Swap x and y

$x = \frac{2y+17}{3y+1}$

Cross Multiply

$(3y + 1)x = 2y + 17$

$3yx + x = 2y + 17$

Subtract x from both sides:

$3yx + x -x= 2y + 17-x$

$3yx = 2y + 17-x$

Subtract 2y from both sides

$3yx-2y =17-x$

Factorize:

$y(3x-2) =17-x$

Make y the subject

$y = \frac{17 - x}{3x - 2}$

Replace y with h'(x)

$h'(x) = \frac{17 - x}{3x - 2}$

9:

$h(c) = \sqrt{2c + 2}$

Replace h(c) with y

$y = \sqrt{2c + 2}$

Swap positions of y and c

$c = \sqrt{2y + 2}$

Square both sides

$c^2 = 2y + 2$

Subtract 2 from both sides

$c^2 - 2= 2y$

Make y the subject

$y = \frac{c^2 - 2}{2}$

$h'(c) = \frac{c^2 - 2}{2}$

10:

$f(x) = \frac{x + 10}{9x - 1}$

Replace f(x) with y

$y = \frac{x + 10}{9x - 1}$

Swap positions of x and y

$x = \frac{y + 10}{9y - 1}$

Cross Multiply

$x(9y - 1) = y + 10$

$9xy - x = y + 10$

Subtract y from both sides

$9xy - y - x = y - y+ 10$

$9xy - y - x = 10$

Add x to both sides

$9xy - y - x + x= 10 + x$

$9xy - y = 10 + x$

Factorize

$y(9x - 1) = 10 + x$

Make y the subject

$y = \frac{10 + x}{9x - 1}$

Replace y with f'(x)

$f'(x) = \frac{10 + x}{9x -1}$

8 0

3 years ago

Pew research reported in 2013 that 15% of American adults do not use the internet or e-mail. They report a margin of error of 2.

Elden [556K]

Answer:

B) They estimate that 2.3% of those surveyed answered incorrectly

Step-by-step explanation:

hope it helps you dude

7 0

3 years ago