Simplify 5 to the tenth power over 5 to the sixth power

I need help please I don’t understand this

mario62 [17]

Answer:

cosB = 21/75, tanB = 72/21

Step-by-step explanation:

Here, we can use pythagoras to figure out all sides. 75^2-72^2 = 21^2. So, we have sides with lengths 21, 72, and 75. We are given that sinB = 72/75. The definition of sin is opposite of hypotenuse. So, the opposite of angle B is side CA and the hypotenuse is BA. This allows us to find that BC is the 21 length side. The cos of an angle is adjacent over hypotenuse. In this case, it is BC/BA which is 21/75. similarly, Tan is opposite over adjacent or 72/21. And we are done. Those are the values. TO remember these, use the acronym:

SOH CAH TOA. S refers to sin and OH refers to opposite over hypotenuse. Similarly C is cos, A is adjacent, and H is hypotenuse, and then T is tan, o is opposite and A is adjacent.

7 0

3 years ago

g The fencing of the left border costs $10 per foot, while the fencing of the lower border costs $2 per foot. (No fencing is req

klio [65]

Answer:

$Area = 1690ft^2\\x = 130ft, y = 26ft$

Step-by-step explanation:

Given the costs we can form an equation:

$10y + 2x =520 - Eq(A)$

and fencing is triangular such that the area enclosed can be written as:

$A = \dfrac{xy}{2} -Eq(B)$

First need to convert the above equation so that it is only in terms of one variable. [either x or y]

To make the equation only in terms of $x$ we can substitute $y$ from Eq(A) i.e, $y =\frac{520-2x}{10}$ , to Eq(B)

$A = \dfrac{x}{2} \dfrac{520-2x}{10}$

simplify

$A = \dfrac{520x - 2x^2}{20}$

$A = -\dfrac{1}{10}x^2 + 26x$

Now, in order to find the maximum area enclosed we can find $\frac{dA}{dx}$ and equate it zero.

$\dfrac{dA}{dx} = -\dfrac{1}{5}x + 26$

$0 = -\dfrac{1}{5}x + 26$

$-26 = -\dfrac{1}{5}x$

$x = 130$

we have the length of one dimension: specifically, the lower fence will be $x =130ft$

we can use this value of $x$ to find the corresponding value of $y$ . From Eq(A)

$10y + 2x =520$

$10y +2(130) = 520$

$y = \dfrac{520 - 2(130)}{10}$

$y = 26$

the length of the left fence will be $y =26ft$

The enclosed area by the fence will be

$A = \dfrac{xy}{2}$

$A = \dfrac{(130)(26)}{2}$

$A = 1690$

Hence the maximum area that can enclosed by the fences provided the costs will be $1690ft^2$

You can even check the cost of the dimensions whether they all add up to $520 or not.

Use Eq(A)

$10y + 2x =520$

$10(26) + 2(130) =520$

and indeed it does!

5 0

3 years ago

Can someone please fill this worksheet out for me? Ive been trying at it for an hour. <br> Thanks!

Tasya [4]

See attached picture for answers:

5 0

3 years ago

A right circular cone has a radius of 6 inches and a volume of 180 cubic inches. What is the height of the cone? A. 5 in. B. 15

dlinn [17]

Answer:

A. 5 in.

Step-by-step explanation:

The formula for the volume of a cone is :

$V= \pi r^{2} \frac{h}{3}$

so

$180 in.^{3} = \pi * 6^{2}* \frac{h}{3}$

$180 in.^{3} = 3.14* 36* \frac{h}{3} \\180 in.^{3} = 113.04 * \frac{h}{3}\\3 * 180 in.^{3} = 113.04 * \frac{h}{3} * 3 \\\\\\\$

$540 / 113.04 = 113.04 h /113.04\\h = 4.8 in$

When you round you get

h= a. 5 in.

6 0

2 years ago

Find the height of the triangle by applying formulas for the area of a triangle and your knowledge about triangles.

neonofarm [45]

Using the Pythagorean Theorem, the altitude to side c has a length of 11.3 inches.

<h3>What is the Pythagorean Theorem?</h3>

The Pythagorean Theorem relates the length of the legs $l_1$ and $l_2$ of a right triangle with the length of the hypotenuse h, according to the following equation: