One of the more familiar squares is 144; the sqrt of 144 is 12. The next higher perfect square is 169, whose square root is 13. Thus, the sqrt of 147 lies between 144 and 169.

7 0

3 years ago

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If an object is dropped from a height of 144 feet, the function h(t)=-16t^2+144 gives the height of the object after t seconds.

shtirl [24]

Answer:

A. 3s

Step-by-step explanation:

The height of the object after t seconds is given by:

$h(t) = -16t^2 + 144$

When will the object hit the ground?

It hits the ground after t seconds, and t is found when $h(t) = 0$ . So

$h(t) = -16t^2 + 144$

$0 = -16t^2 + 144$

$16t^2 = 144$

$t^2 = \frac{144}{16}$

$t^2 = 9$

$t = \sqrt{9}$

$t = 3$

3 seconds, so the correct answer is given by option a.

5 0

3 years ago

Find the geometric mean of 11 and 23. Round your answer to the nearest tenth.

djyliett [7]

Answer: 15

Step-by-step explanation:

4 0

3 years ago

during a softball game kay hit a fly ball the function f(x) = -16t^2 + 64t + 4 describes the height of the softball in feet. mak

STatiana [176]

<h2>Hello!</h2>

The graph is attached.

To graph a parabola we need to know the following:

- If the parabola is open upwards or downwards

- They axis intercepts (if they exist)

- The vertex position (point)

We are given the function:

$f(t)=-16t^{2}+64t+4$

Where,

$a=-16\\b=64\\c=4$

For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.

Finding the axis interception points:

Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:

$f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0$

Using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615$

So, at t=4.0615 the height of the softball will be 0.

Since we will work only with positive values of "x", since we are working with a function of time:

Let's start from "t" equals to 0 to "t" equals to 4.0615.

So, evaluating we have:

$f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0$

Finally, we can conclude that:

- The softball reach its maximum height at t equals to 2. (68 feet)

- The softball hits the ground at t equals to 4.0615 (0 feet)

- At t equals to 0, the height of the softball is equal to 4 feet.

See the attached image for the graphic.

Have a nice day!

3 0

3 years ago