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3 years ago

6

Which pairs of angles in the figure below are vertical angles? Check all that apply.

2 answers:

salantis [7]3 years ago

7 0

Answer:

C Only

Step-by-step explanation:

They are the only pair of vertical angles.

photoshop1234 [79]3 years ago

5 0

Answer: the answers are C and D because they are across from each other

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If 0 ≤ Ø < 90 degrees, and tan Ø= 2/3, what is sinØ + cosØ?

Alika [10]

Answer:

1.38

Step-by-step explanation:

tan Ø = 2/3

Ø = tan⁻¹(2/3)

Ø = 33.69

sin 33.69 ≈ 0.55

cos 33.69 ≈ 0.83

sin Ø + cos Ø ≈ 1.38

4 0

3 years ago

X2+4x+3=0 using quadratic formula

Elenna [48]

The correct answer is x= -1 or -3

6 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

2 years ago

Help me please? thanks so much.

salantis [7]

1 is C
2 is A

Those are the answers

7 0

2 years ago

140 students are going on a field trip they fill up a total of 4 busesequally how many students are in each bus

Nana76 [90]

Answer:

560

Step-by-step explanation:

8 0

3 years ago