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3 years ago

Andre correctly solved an equation using the steps shown:

Mathematics

1 answer:

Pepsi [2]3 years ago

3 0

Answer:

This equation has zero solutions.

Step-by-step explanation:

This equation has zero solutions because 2 is not equal to zero.

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A gift box is in the shape of a right rectangular prism, as pictured below.What is the volume, In cubic centimeters, of the gift

nevsk [136]

The length is 24 centimeters and the width is 10 centimeters and the height is 13 centimeters

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3 years ago

What are two different ways you could find the value of a? Explain these methods.

tekilochka [14]

Step-by-step explanation:

What are two different ways you could find the value of a? Explain these methods. A right triangle is shown. An altitude is drawn from the right angle to the opposite side to form 2 line segments with lengths 9 and 16. The length of the other 2 sides are 15 and a....done

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2 years ago

Sin 3pi divided by 4

KATRIN_1 [288]

Sin 3pi/4 is angle 135 degrees or 45 degrees below 180 degrees. Hence it's opposite side is 1 and adjacent is 1, implying the hypothenus is sqrt(2). Hence

Sin 3pi/4 = 1/sqrt(2). Now multiply by sqrt(2)/sqrt(2) to get:

[1/sqrt(2).] * [sqrt(2)/sqrt(2)] = sqrt2)/2

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3 years ago

Can someone please help me?

Crazy boy [7]

Answer:

70°

Step-by-step explanation:

since two lengths are equal to 3.5 that’s means there’s are two equal angles (55° and 55°)

a triangle adds up to 180°

so to find x we set this up:

x +55 +55 = 180

x + 110 = 180

x = 70°

3 0

3 years ago

I am offering another 100 points

Ivahew [28]

Answer:

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

Step-by-step explanation:

Perfect squares: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find $\sf \sqrt{75}$ , identify the perfect squares immediately before and after 75:

64 and 81

$\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}$

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

See the attachment for the correct placement of $\sf \sqrt{75}$ on the number line.

6 0

11 months ago

Read 2 more answers