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3 years ago

What is the value of f(5) ?

Mathematics

1 answer:

qwelly [4]3 years ago

6 0

Answer:

243

Step-by-step explanation:

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3 years ago

Use the triangle below to find cos V.

Semenov [28]

Answer:

$cos\ V = \frac{2\sqrt{29}}{29}$

Step-by-step explanation:

Given

The attached triangle

Required

Find cos V

In trigonometry:

$cos\theta = \frac{Adjacent}{Hypotenuse}$

In this case:

$cos V= \frac{TV}{UV}$

Where

$TV = 6$

and

$UV^2 = TV^2 + TU^2$ -- Pythagoras

$UV^2 = 6^2 + 15^2$

$UV^2 = 261$

Take square roots

$UV = \sqrt{261$

$UV = \sqrt{9*29$

$UV = \sqrt{9} *\sqrt{29$

$UV = 3\sqrt{29$

So:

$cos V= \frac{TV}{UV}$

$cos\ V = \frac{6}{3\sqrt{29}}$

$cos\ V = \frac{2}{\sqrt{29}}$

Rationalize:

$cos\ V = \frac{2}{\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}$

$cos\ V = \frac{2\sqrt{29}}{29}$

8 0

3 years ago

A closed box has a square base with side length feet and height feet given that the volume of the box is 34 ft.³ express the sur

Deffense [45]

Answer:

$\dfrac{2s^3+136}{s}$

Step-by-step explanation:

Let the side length of the square base =s feet

Let the height of the box = h

Given that the volume of the box = $34$ ft^3$

Volume of the box = $s^2h$

Then:

$s^2h=34$ ft^3\\$Divide both sides by s^2\\h=\dfrac{34}{s^2}$

Surface Area of a Rectangular Prism =2(lb+bh+lh)

Since we have a square base, l=b=s feet

Therefore:

Surface Area of our closed box $= 2(s^2+sh+sh)$

$S$urface Area= 2s^2+4sh\\Recall: h=\dfrac{34}{s^2}\\$Surface Area= 2s^2+4s\left(\dfrac{34}{s^2}\right)\\=2s^2+\dfrac{136}{s}\\$Surface Area in terms of length only=\dfrac{2s^3+136}{s}$

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3 years ago