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ioda
3 years ago
15

A dripping faucet leaps 1.8 ounces per minute. How many ounces leak in 30 minutes?

Mathematics
1 answer:
zzz [600]3 years ago
5 0

Answer:

54 ounces

Step-by-step explanation:

1.8 * 30 = 54

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A statue is mounted on top of a 16 foot hill. From the base of the hill to where you are standing is 77 feet and the statue subt
DanielleElmas [232]

Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.

You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:

\tan 13.2^{\circ}=\dfrac{\text{opposite leg}}{\text{adjacent leg}} =\dfrac{16+h}{77}, \\ \\ 0.2345=\dfrac{16+h}{77},\\  \\ 16+h=0.2345\cdot 77=18.0565,\\ h=18.0565-16=2.0565 ft.

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3 years ago
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A type of sampling in which individuals to be interviewed are selected based on their proportion or quota in the general populat
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A type of sampling in which individuals to be interviewed are selected based on their proportion or quota in the general population being polled is known as public opinion.

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6 0
2 years ago
What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
4 years ago
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