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zlopas [31]
2 years ago
15

HELP say im getting it wrong the perimeter of the polygons is ?​

Mathematics
1 answer:
lara [203]2 years ago
8 0
So the answer would be16*4=64
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Use the distributive property to rewrite (5x + 3)(2) as a sum of two terms. <br><br> please hurry
Anna007 [38]

Answer:

10x + 6

Step-by-step explanation:

When distributing, you need to multiply the number on its own by both terms of the binomial (the 5x + 3). 5 times 2 is 10, and then you have the x, so it is 10x. Multiply 3 times 2 and get 6. Hope this helps!

6 0
3 years ago
How much would it cost him in interest over 5 months if he used card A to make the purchase?
Kamila [148]

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I need a little bit more detail in the question to answer.

Step-by-step explanation:

4 0
2 years ago
Which figure is NOT a polygon?
alexdok [17]

Answer:

A.

Hope It Help

Step-by-step explanation:

Brainliest please

5 0
2 years ago
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A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x &lt; 0 and p(x) = 1 for x &gt
34kurt
F_X(x)=\begin{cases}0&\text{for }x2\end{cases}

Recall that the PDF is given by the derivative of the CDF:

f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}

The mean is given by

\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23

The median is the number M such that F_X(x)=\mathbb P(X\le M)=\dfrac12. We have

F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2

but both roots can't be medians. As a matter of fact, the median must satisfy 0\le M\le2, so we take the solution with the negative root. So M=2-\sqrt2 is the median.
3 0
3 years ago
How many numbers can you make with four digits
Natasha_Volkova [10]
Well, first consider how many digits there are. You have 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0, depending on how you use it, that are all digits.
If you count 0 as a digit other than a place holder (0000, 0001, etc.), you would end up with ten thousand (10,000) numbers with four digits, starting with 0000, and ending with 9999. If you count 0 as nothing other than a place holder (1000, 1001, etc.), you would have nine thousand (9,000) numbers with four digits, starting with 1000.
So, depending on how you view 0, you can make up to 10,000 different numbers that contain four digits.
5 0
3 years ago
Read 2 more answers
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