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3 years ago

HELPPPPP ASAPPPP

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

The maximum height of the volleyball is 12.25 feet.

Step-by-step explanation:

The height of the volleyball, h(t), is modeled by this equation, where t represents the time, in seconds, after that ball was set :

$h(t)=-16t^2+20t+6$ ....(1)

The volleyball reaches its maximum height after 0.625 seconds.

For maximum height,

Put $\dfrac{dh}{dt}=0$

Now put t = 0.625 in equation (1)

$h(t)=-16(0.625)^2+20(0.625)+6\\\\h(t)=12.25\ \text{feet}$

So, the maximum height of the volleyball is 12.25 feet.

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AnnZ [28]

Answer:

D. Obtuse.

Step-by-step explanation:

We are told that in $\Delta MNP$ $m\angle M=35^o$ and $m\angle P=47^o$ . We are asked to classify our given $\Delta MNP$ .

We can see that angle M and angle P are acute angles as their measure is less than 90 degrees.

Let us find the measure of angle P using angle sum property of triangles, which states that sum of interior angles of a triangle is 180 degrees.

So we can set an equation as:

$m\angle M+m\angle P+m\angle N=180^o$

Upon substituting our given values we will get,

$35^o+47^o+m\angle N=180^o$

$82^o+m\angle N=180^o$

$82^o-82^o+m\angle N=180^o-82^o$

$m\angle N=98^o$

As measure of angle N is 98 degrees, so angle N is an obtuse angle.

Since a triangle having an angle that measures more than 90 degrees is called an obtuse triangle, therefore, $\Delta MNP$ is an obtuse triangle and option D is the correct choice.

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3 years ago

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so

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Answer:

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