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2 years ago

A pogo stick bounces for 30 m in 10 S what is the bouncers average speed

Mathematics

1 answer:

kenny6666 [7]2 years ago

3 0

Answer:

3m/s

Step-by-step explanation:

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Ms. Ross has 24 students. 2/3 of her students have brown hair. The rest have blonde. How many students have brown?

Dimas [21]

The answer would be 16 because 24 divided by 3 is 8. 8 multiplied by 2 is 16. 16 students have brown hair.

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3 years ago

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June has 30 days. A family of 3 used 3,960 gallons of water during the month of June. How many gallons of water did each person

lys-0071 [83]

Answer:

44 gallons

Step-by-step explanation:

3960/3=1320

Each person used 1320 in one month

1320/30=44

Amount of person used / 30 days = gallons used each day

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3 years ago

Which pair of angles is an example of supplementary angles?

artcher [175]

Answer:

A. Angle 1 and Angle 7

Step-by-step explanation:

Those two are the only angles that when added up, the sum is 180.

Angle 7 and Angle 2 are also congruent, so if Angle two forms a straight line with Angle 1, Angle 7 should be able to do that too.

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2 years ago

PLEASE ANSWER DUE IN 30 MINUTES <br> I WILL MARK BRAINLIEST<br> QUESTIONS 1&2 AND SHOW WORK

zhuklara [117]

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Step-by-step explanation:

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2 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago