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3 years ago

Which equation represents a function?

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

All the answers except C has a Vertical Line when graphed. It must pass the Vertical line test in order to be a function.

Vertical Line Test: If you can draw a vertical line anywhere on a graph so that it hits the graph in more than one spot, then the graph is NOT a function.

C.) can pass the test since it's a horizontal line.

Hope this helps!

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Please help. I’ll mark you as brainliest if correct

zysi [14]

Answer:

12 + -6i

a=12

b=-6

Step-by-step explanation:

( -4 + 3i ) ( -3 - 2i )

-4 * -3 = 12

3i * -2i= -6i

12 + -6i

6 0

3 years ago

5y + 4(-5 + 3y) = 1 – y solve step by step please

Studentka2010 [4]

Step by step, meaning explaining the equation right and answer?

3 0

3 years ago

Read 2 more answers

One week a business sold 40 shirts. White ones cost $4.95, and the printed ones cost 7.95. In al $282 worth of shirts were sold.

Tanya [424]

Answer:

12 white 28 printed

Step-by-step explanation:

x = white y = printed

x + y = 40————(1)

4.95x + 7.95y = 282————(2)

Solve for x using 1st equation

x = 40 - y Plug in for x using 2nd equation

4.95(40 - y) + 7.95y = 282 Solve for y

198 - 4.95y + 7.95y = 282

3y = 84

y = 28 Plug in for y using 1st equation

x + (28) = 40 Solve for x

x = 12

I hope this helped and please mark me as brainliest!

8 0

2 years ago

Read 2 more answers

Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.

skad [1K]

DfLet $f(x)=\sec x$ and $g(x)=\sqrt x$ . Then

$y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)$

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

where $u=g(x)=\sqrt x$ , so that $y=f(g(x))=f(u)=\sec u$ . We have

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}$
$\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}$

5 0

3 years ago

Compound Inequalities <br><br> (Help Please)

erastova [34]

2x + 3 ≥ 7
2x ≥ 7 - 3
2x ≥ 4
x ≥ 4 / 2
x ≥ 2

2x + 9 > 11
2x > 11 - 9
2x > 2
x > 1

Those are your two inequalities, now we must find one that satisfies both. Combine them together you get:

1 < 2 ≤ x

That last inequality becomes your answer.

x ≥ 2 makes both statements true.

3 0

3 years ago