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3 years ago

14

A professional volleyball player can jump an average of 2.5 times higher than the average person. if an average person can jump

inches high, how high can the professional volleyball player jump?

1 answer:

nikitadnepr [17]3 years ago

5 0

Lets say average person jumps A inches high, then volleyball player can jump 2,5 x A inches high.

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nadezda [96]

Answer:

(-6,4)

Step-by-step explanation:

If this is the system of equations here is how:

solve for 1 vairuble in the first equation, and plug that it to the other.

3 0

3 years ago

Y=4mx-7b solve for x

muminat

Your answer is x = 7b/4m + y/4m

8 0

4 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0

3 years ago

The perimeter of a parallelogram is 72 meters. The width of the parallelogram is 4 m less than its length. find the link and the

AlladinOne [14]

The width is 16 and the length is 20

8 0

3 years ago

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve

sweet [91]

A conservative vector field $\mathbf f$ has curl $\nabla\times\mathbf f=\mathbf0$ . In this case,

$\nabla\times\mathbf f=12xy^2z(1-z)\,\mathbf j+4yz^2(2z-3x)\,\mathbf k\neq\mathbf 0$

so the vector field is not conservative.

3 0

3 years ago