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ludmilkaskok [199]

2 years ago

9

Juan is rappelling down a 495-foot cliff at a rate of 6 feet per second. Which equation would represent the height Juan is above

the ground, H, after t seconds?

1 answer:

ss7ja [257]2 years ago

8 0

Answer:

The equation is H = 495 -6t

Step-by-step explanation:

Here, we want to write an equation that will represent the height Juan is above the ground, after t seconds.

The height we want to calculate is H. The initial height is 495 foot;

At the first second, the decrease in height will be 6 * 1, at the second 6 * 2 and thus at t seconds, the decrease in height will be 6 * t = 6t

Ask the height after t seconds will be;

H = 495 - 6t

Where t represents the time in seconds

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Answer needed ASAP please!!

barxatty [35]

Answer:

As can be clearly seen from the attached diagram angles PBC and BAD are congruent by the Corresponding Angles Theorem AB acts as the transversal for the lines BC and AD which are parallel.

Also, it can be clearly seen that the angles ABC and BAT are congruent by the Alternate Interior Angles Theorem as AB acts as the transversal for the lines BC and TAD which are parallel.

Thus, the only option from the given list of options which matches the answer is Option A.

Thus, Option A is the correct option and thus the final answer.

Step-by-step explanation:

5 0

2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

If a gallon has 4 quarts, and 1 has quart has 2 pints, how many pints are in 8 gallons

Soloha48 [4]

Answer:

I also think 62 correct me if I'm wrong

3 0

2 years ago

a bathtub is filled with 110 gallons of water. The tub drains at a rate of 5 gallons every 2 minutes. How long will it take befo

Gnesinka [82]

Answer:

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2 years ago

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Marina CMI [18]

Answer:

The answer is soooo simple

Step-by-step explanation:

its c

4 0

2 years ago