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3 years ago

Which set of ordered pairs represents y as a function of x?

Mathematics

1 answer:

stiks02 [169]3 years ago

4 0

Answer:

y = 4x, y = 10x, etc.

Step-by-step explanation:

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Benito is selling T-shirts for $8 each for his school fund-raiser. So far, he has sold 16 T-shirts. How many more does he need t

zhannawk [14.2K]

D. When you do the math, the results to everything should be t=9. D is the only outlier. Therefore, D is the answer in this case.

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4 years ago

How do I get the answer for exponents? Please explain step by step to get marked!

ioda

Answer:

4900

Step-by-step explanation:

70^2 means 70*70

70*70=4900

3 0

3 years ago

Read 2 more answers

Which function does not have a vertical asymptote

Gemiola [76]

Answer:

$y=\frac{5x}{1+2x^2}$

Step-by-step explanation:

$Option ~A ~ does ~ not ~ have ~a ~ vertical~ asymptote$

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Hope it helps...

Have a great day!!

8 0

3 years ago

A bat and a ball cost $1.10. The bat costs one dollar more than the ball. How much does the ball cost? Correct Answer Gets Brain

KatRina [158]

Answer:

0.10

Step-by-step explanation:

If together the ball and bat cost $1.10 and the bat cost one more dollar than the ball than you would have too subtract $1 from what the ball and bat together. so 0.10 would be left

8 0

3 years ago

Read 2 more answers

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

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3 years ago