State if the two triangles are congruent. If they are, state how you know.

A rectangular yard measuring 39 ft by 67 ft is bordered (and surrounded) by a fence. Inside, a walk that is 4 ft wide goes all t

Gekata [30.6K]

The area of the walk is 784 square feet

Area is the quantity that expresses the extent of a region on the plane or on a curved surface

First, calculate the area of the rectangular yard.

Area = L x W

where A equals area, L equals length, and W equals width.

According to the stated problem:

A = 39 x 67

A = 2613 square feet <-------Area of Rectangular yard.

Now, according to the stated problem there is a walk that is 4 feet wide going all the way along the fence (Inside). "

Therefore, we have 8 feet to subtract from each of the values above (39) and (67).

That is 4 feet on each side,

A = L x W

A = 31 x 59

A = 1829 square feet <----------Area of new rectangle formed by allowing for three foot walk.

Now, subtract the area of the new rectangle formed from the area of the original area. Doing so, will give you the area of the walk, correct?

Original area = 2613 square feet

New area = 1829 square feet

Hence, the Area of WALK = 784 square feet

Learn more about area on: brainly.com/question/22972014

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3 0

1 year ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

3m-5p=12 solve for p

umka21 [38]

3m - 5p = 12

3m(-3m) -5p = (-3m)+ 12
-5p = -3m + 12
(-5)/-5p = (-5)/ (-3m + 12)
p = 3/5m + 12/-5

5 0

3 years ago

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Please Help!!!!!!!!!

marissa [1.9K]

124 square units
you just have to count the top, bottom, and a side and times by 2

4 0

3 years ago

2. Three square rooms in Joey's house meet to form a right triangle as shown. Joey

Oxana [17]

Answer:

The amount of carpet needed in Room Z should be the same as the carpet needed in both Room X and Room Y.

Step-by-step explanatio

3 0

3 years ago

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