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3 years ago

Can you help me with this?

Mathematics

1 answer:

satela [25.4K]3 years ago

8 0

I cant hardly see it ,what's the question?

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3 years ago

Find the volume of the cone

kompoz [17]

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V≈628.32 units³

Step-by-step explanation:

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V=πr2h/3 (h/3 as in a fraction)

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3 years ago

Read 2 more answers

Let f(x) = twenty-four over the quantity of 3 x minus two. Find f(-2

bija089 [108]

$f(x)=\frac{24}{3x-2}\\\\f(-2)=?\\
Subtitude\ x=-2\\
f(-2)=\frac{24}{3*(-2)-2}=\frac{24}{-6-2}=\frac{24}{-8}=-3\\\\
Solution\ is\ f(-2)=-3.$

4 0

3 years ago

Read 2 more answers

zmey [24]

Answer:

Part 1) The inequality that represent this situation is $5(x+7)^{2} \geq1,050$ or $5x^{2}+70x+245 \geq1,050$

Part 2) Yes, 8 inches is a reasonable width for his tablet

Step-by-step explanation:

Part 1)

Let

L -----> the length of the screen television

W ----> the width of the screen television

x ----> the width of Andrew's tablet

we know that

$L=5W$ ------> equation A

$W=x+7$ ----> equation B

The area of the television is

$A=LW$ -----> equation C

Substitute equation A and equation B in equation C

$A=5(x+7)(x+7)$

$A=5(x+7)^{2}$

$5(x+7)^{2} \geq1,050$

$5(x^{2}+14x+49) \geq1,050$

$5x^{2}+70x+245 \geq1,050$ ------> inequality that represent this situation

Part 2) Determine if 8 inches is a reasonable width for his tablet

For x=8 in

Substitute in the inequality

$5(8+7)^{2} \geq1,050$

$5(15)^{2} \geq1,050$

$1,125 \geq1,050$ -----> is true

therefore

Yes, 8 inches is a reasonable width for his tablet

3 0

3 years ago

Solve for x. Round your answer to the nearest tenth.

Rainbow [258]

Answer:

24.8

Step-by-step explanation:

x = √(16)^2+(19)^2= 24.8

7 0

2 years ago