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3 years ago

Percy is filling a cylindrical tank with water. The tank is

Mathematics

2 answers:

Archy [21]3 years ago

7 0

Answer:

Step-by-step explanation:

78x9=702

702/6=117 minutes.

Ber [7]3 years ago

6 0

Volume of water in the tank:

Differentiate both sides with respect to time t :

V changes at a rate of 2000 cc/min (cubic cm per minute); use this to solve for dh/dt :

(The question asks how the height changes at the exact moment the height is 50 cm, but this info is a red herring because the rate of change is constant.)

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40 POINTS PLZ ANSWER PLZ HELP

Tanya [424]

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Which of the following is the least? A. 0.105 B. 0.501 C. 0.015 D. 0.15

Naddika [18.5K]

C is the least that is beause it is .015 which is smaller than 0.15

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3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

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4 years ago

Plz its easy :) plzz

Thepotemich [5.8K]

Here is the answer hope it helps

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4 years ago

Read 2 more answers

Combining like terms in a quadratic expression.... simplify -11x^2 - 3x - 7x^2 + 9 + 8x HELP PLEASE! AND A FEW OTHERS!

valentina_108 [34]

The answer is 4x^2 +5x+9

You will first solve the same variables such as x^2

11x^2 - 7x^2 = 4x^2

now you will solve the variables with x

-3x +8x = 5x

so the answer you will get will be 4x^2 +5x = 9

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3 years ago