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3 years ago

I need help. Of the make sense of the problem section. Please

Mathematics

1 answer:

uranmaximum [27]3 years ago

5 0

the basket balls height when bounced can vary. depending on the strength of ball being thrown to the ground can make the ball bounce even higher. So for an example, if you applied full force from a average weighted male adult. The basket ball will bounce a decent height. But, of course will for sure not be as high as to the first bounce.

hope this helps

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Plsss helppppp I’ll mark u brainlist no links pls

Basile [38]

Step-by-step explanation:

you flip it from the line, match the letters (Z woube be C, X would be A, etc.) and then writw the new coordinates

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3 years ago

A snowball has a radius of 7cm. What is the volume of the snowball?

BabaBlast [244]

Answer:

The volume is 1436.8 $in^2$ .

Step-by-step explanation:

-To find the volume of a sphere, you need the formula:

$V = \frac{4}{3} \pi r^3$

-Use the radius for the formula:

$V = \frac{4}{3} \pi 7^3$

-Then, you solve:

$V = \frac{4}{3} \pi 7^3$

$V = \frac{4}{3}$ × $\pi$ × $7^3$

$V = \frac{4}{3} \pi$ × $343$

$V = \frac{1372}{3} \pi = 1436.755$

-Round the answer to the nearest tenth:

$V = 1436.8$

So, therefore the volume is 1436.8 $in^2$ .

7 0

3 years ago

We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure

STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be $h(t) = -16\cdot t^{2} + 128\cdot t + 320$ , the first and second derivatives are, respectively:

First Derivative

$h'(t) = -32\cdot t +128$

Second Derivative

$h''(t) = -32$

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

$-32\cdot t +128 = 0$

$t = \frac{128}{32}\,s$

$t = 4\,s$ (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

$h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320$