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Svetlanka [38]
3 years ago
7

Find the perimeter of shape #8

Mathematics
1 answer:
Wittaler [7]3 years ago
4 0
The squiggly lines and the 2 straight lines through to sides of the pentagon mean that they are congruent (equal in length).

So, to find the perimeter of a regular polygon, you add all of its sides together.


So, the perimeter= 4x-2+x-2+x-2+x+2+x+2

To simplify it, add all of the x's together.

4x+x+x+x+x=8x

Perimeter=8x-2-2-2+2+2

-2-2=-4-2=-6+2=-4+2=-2

Perimeter=8x-2

Reach out to me if you have any questions. I'd be glad to help! :)
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Place Math Quiz
Aleksandr [31]

Answer:

c) $136,800

Step-by-step explanation:

Since the thing is 3 years and in 3 year there are 36 months. And since $3,800/mo so 36 x 3,800 = 136,800

So the answer is $136,800

6 0
3 years ago
A small greeting card company prints and ships greeting cards throughout the year. Rodney is able to package and ship 416 cards
Rasek [7]

Answer:

52 cards/hour

Step-by-step explanation:  

                                                        416 cards

Rodney's card production rate is --------------- = 52 cards/hour

                                                          8 hours

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E3_t%7B%7Cx%2B2%7C%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limi
photoshop1234 [79]
\displaystyle\int_{-3}^3|x+2|\,\mathrm dx

Recall the definition of absolute value:

|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x

So we can split up the integration interval at x=-2 and apply this definition to rewrite the integral as

\displaystyle\int_{-3}^{-2}(-(x+2))\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx
=\displaystyle-\int_{-3}^{-2}(x+2)\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx
=-\left(\dfrac12x^2+2x\right)\bigg|_{x=-3}^{x=-2}+\left(\dfrac12x^2+2x\right)\bigg|_{x=-2}^{x=3}
=\dfrac12+\dfrac{25}2=13
7 0
3 years ago
You have 128 balloons to use for your party. You want to tie them to 16 posts. How many balloons can you put on each post?
marta [7]
Divide 128 by 16 to get how many balloons will go on each post.

128 ÷ 16 = 8

8 balloons per post :)
3 0
3 years ago
Let f(x) = (x^2+3x-5)/(x^2-8x+15) What is the domain of f(x) ?
Lyrx [107]

Answer:

So, Domain = {set of all the real numbers except 3, 5}

Interval notation = {x ∈ set of all real numbers/ x ≠3 and x ≠ 5}

Step-by-step explanation:

The given function is f(x) = \frac{x^2 + 3x - 5}{x^2 - 8x + 15}

Here we need to find the domain.

To find the domain, let's find the restricted domain.

If we set the denominator equal to zero and solve for x, we can find the restricted domain.

Here the denominator is x^2 -8x + 15 = 0

Now we can factorize and solve for x.

(x - 3)(x - 5) = 0

(x - 3) or (x - 5) = 0

x -3 = 0  and x - 5 = 0

Solving the above equations, we get

x = 3 and x = 5

The restricted domains are x = 3 and x = 5.

Now we can find the domain.

Except 3 and 5, all the real numbers can be domain of this function.

So, Domain = {set of all the real numbers except 3, 5}

= {x ∈ set of all real numbers/ x ≠3 and x ≠ 5}.

8 0
3 years ago
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