Ask question

Ask question

All categories

3 years ago

14

PLEASE HELP ASAP!!!!!!!!!!!!!

1 answer:

Step2247 [10]3 years ago

5 0

Answer

12. x = 15, y = 21

13. x = 10, M = 37, N = 84, P =59

14. x = 19, R = 43, S = 20, T = 117

15. A = 21, B = 125, C = 34

16. J = 48, K = 57, L = 75

Explanation

12. 6x - 11 + x + 5 + 81 = 180

7x + 75 = 180

7x = 105

x = 15

4y - 18 + y + 14 + 6*15 - 11 = 180

5y + 75 = 180

5y = 105

y = 21

13. 4x - 3 + 9x - 6 + 6x - 1 = 180

19x - 10 = 180

19x = 190

x = 10

M = 4*10 - 3 = 37

N = 9*10 - 6 = 84

P = 6*10 - 1 = 59

14. R = 2x + 5

S = x + 1

T = 7x - 16

2x + 5 + x + 1 + 7x - 16 = 180

10x - 10 = 180

10x = 190

x = 19

R = 2*19 + 5 = 43

S = 19 + 1 = 20

T = 7*19 -16

15. A = C -13

B = 4C - 11

c - 13 + 4c - 11 + c = 180

6c -24 = 180

6c = 2-4

C = 34

A = 34 - 13 = 21

B = 4*34 - 11 = 125

16. K = J + 9

L = 2J -21

j + 9 + 2j - 21 + j = 180

4j -12 = 180

4j = 192

J = 48

K = 48 + 9 = 57

L = 2*48 -21 = 75

You might be interested in

When Benny’s sister penny is 24 Benny’s age will be 125% of her age. A) How old is Benny?

vlada-n [284]

A) Benny is 30 years old

B) I think Penny isn’t born yet

3 0

3 years ago

Question number 2 please ..

fredd [130]

Nz = 14

I think that is the answer

7 0

3 years ago

What is 1 2/10+8 1/18=

n200080 [17]

Answer: ${9}\frac{23}{90}$

Step-by-step explanation:

5 0

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

Find the linear equation that is the straight line through (1, 2) and (8, 30).

marissa [1.9K]

Slope
= 30-2/8-1

= 28/7

= 4

Equation:
y - 2/x - 1 = 4

y - 2 = 4(x-1)

y - 2 = 4x - 4

y = 4x - 4 + 2

y = 4x - 2

3 0

2 years ago

Read 2 more answers