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3 years ago

I need help with another set of math questions.

Mathematics

1 answer:

Anit [1.1K]3 years ago

7 0

For question 11, you essentially need to find when h(t) = 0, since that is when the height of the ball reaches 0 (ie touches the ground).

For question 12, it is asking for a maximum height, so you need to find when dh/dt = 0 and taking the second derivative to prove that there is maximum at t. That will find you the time at which the ball will hit a maximum height.

Rinse and repeat question 12 for question 13

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Can a 180 degrees angle be a obtuse triangle?

Aleksandr [31]

Answer:

Yes and no, I'll explain below

Step-by-step explanation:

An obtuse triangle can have either a 90 degree angle or an 180 degree angle. This reason being said, an obtuse triangle has ONLY one angle that is greater than 90 degrees. But the sum of any given triangle is 180. Hope this helps!

4 0

2 years ago

A rectangular prism has a length of 5 1/8 feet, a width of 7 1/2 feet, and a height of 2 feet. What is the volume of the prism?

maksim [4K]

For this case we have that by definition, the volume of a rectangular prism is given by:

$V = A_ {b} * h$

Where:

$A_ {b}:$ It is the area of the base

h: It's the height

According to the data we have:

$length = 5 \frac {1} {8} = \frac {8 * 5 + 1} {8} = \frac {41} {8}$

$width = 7 \frac {1} {2} = \frac {2 * 7 + 1} {2} = \frac {15} {2}$

Then:

$A_ {b} = \frac {41} {8} * \frac {15} {2} = \frac {615} {16}$

Thus, the volume is:

$V = \frac {615} {16} * 2 = \frac {1230} {16} = 76.875$

Answer:

$76.875 \ ft ^ 3$

6 0

3 years ago

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$