Write down the formula to calculate the area of an isosceles triangle whose equal sides are 'a' unit and base is 'b'unit.

Mathematics

1 answer:

Elena-2011 [213]3 years ago

7 0

Step-by-step explanation:

I hope this will help you..........

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The formula for the circumference of a circle is c= 2pi r, where r is the radius. Rearrange the formula to solve for our and sel

ratelena [41]

Answer: r = C/(2pi)

This is the same as writing $r = \frac{C}{2\pi}$

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Explanation:

pi is some number (approximately 3.14) which means 2pi is also some number (roughly 6.28)

Saying c = 2pi*r means we have 2pi times r, and the result is the circumference c. To isolate r, we'll undo the multiplication. We'll undo it by dividing both sides by 2pi like so

C = 2pi*r

C/(2pi) = r

r = C/(2pi)

in which we can write it like $r = \frac{C}{2\pi}$

So whatever C is, we divide it over 2pi (aka roughly 6.28) to get the radius.

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Extra Info (Optional Section):

As an example, let's say the circumference of the circle is 628 feet. This means the distance around the circle is 628 feet. So C = 628 would lead to...

r = C/(2pi)

r = C/(6.28)

r = 628/(6.28)

r = 100

So the radius would be roughly 100 feet.

8 0

3 years ago

A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height

melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

$x\geq 0$

The height of the food packet cannot be a negative value, so

$f(x)\geq 0$

We need to replace $-15x^2+6000$ for f(x) in the above inequality to find the domain.

$-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0$

The possible solutions of the above inequality are given by the intervals $(-\infty , -2], [-2,2], [2,\infty )$ . We need to pick test point from each possible solution interval and check whether that test point make the inequality $(x+200)(x-200)\leq 0$ true. Only the test point from the solution interval [-200, 200] make the inequality true.