What is the perimeter of the rectangle

2 answers:

7 0

Well, you would find the perimeter adding up all of the sides. Would you please tell us what each of the sides are, or what sides are what? :)

Pani-rosa [81]4 years ago

6 0

To find the perimeter you have to add all the sides of the rectangle. You can also do P = 2(L) + 2(W)

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A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(

masha68 [24]

For this case we have the following function:
f (t) = -16t2 + 32t + 384
By definition we have to:
The average rate of change:
AVR = (f (t2) - f (t1)) / (t2 - t1)
Where,
For t1 = 4
f (4) = -16 * (4) ^ 2 + 32 * (4) + 384
f (4) = 256
For t2 = 6
f (6) = -16 * (6) ^ 2 + 32 * (6) + 384
f (6) = 0
Substituting values we have:
AVR = (0 - 256) / (6 - 4)
AVR = -128
Answer:
The average rate of change of f (t) from t = 4 seconds to t = 6 seconds is -128 feet per second.

6 0

3 years ago

Read 2 more answers

In many cases the Law of Sines works perfectly well and returns the correct missing values in a non-right triangle. However, in

Agata [3.3K]

$\frac{\sin B}{b}=\frac{\sin A}{a} \\ \\ \sin B=\frac{b \sin A}{a} \\ \\ \sin B=\frac{13 \sin 29^{\circ}}{7.2} \\ \\ \boxed{m\angle B =\arcsin \left(\frac{65\sin 29^{\circ}{36} \right), 180^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$

Considering the diagram. $\angle B$ is shown to be acute, so $\boxed{m\angle B=\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$ .

Angles in a triangle add to 180°, so $m\angle C= 151^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)$ .

Using the law of cosines,

$AB=\sqrt{13^2 + 7.2^2-2(13)(7.2)\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)} \\ \\ \boxed{AB=\sqrt{220.84-187.2\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)}}$