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barxatty [35]
3 years ago
15

The ______ of the degrees of each factor is

Mathematics
2 answers:
nasty-shy [4]3 years ago
8 0

Answer:

  • B. Sum

Step-by-step explanation:

When finding the volume, we multiplied the 3 dimensions with degree of 1, 1 and 3 and got the binominal of degree 5. We summed up the degrees: 1 + 1 + 3 = 5.

  • The <u>SUM</u> of the degrees of each factor is  the degree of the product.
jeyben [28]3 years ago
6 0

Answer:

sum

Step-by-step explanation:

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Name the like terms <br> 5a, 3, 4d , 2a, 2a, 5c
Vsevolod [243]

Answer:

"Like terms" are terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are "like" each other. Note: the coefficients (the numbers you multiply by, such as "5" in 5x) can be different.

Step-by-step explanation: Hope this helps

6 0
2 years ago
14tx=46;x=32 what's the answer?​
pickupchik [31]

First plug 32 in for x 14(32)t=46

Now solve for t

448t=46

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3 years ago
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2(7-3)+4(10÷2) use the oder of operations to simplify ​
MAVERICK [17]

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Step-by-step explanation:

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3 years ago
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Gekata [30.6K]

See the attached image below for the dot plot. The center is at 8 with dots on either side of 8 being a mirror copy of one another. For example, there are 2 dots over 6 and 2 dots over 10

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3 years ago
Find the general solution of <img src="https://tex.z-dn.net/?f=2cos%282x%20%2B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%29%20%3D%20%5Cs
Aleks04 [339]

Answer:

\large\boxed{x=-\dfrac{\pi}{4}+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}

Step-by-step explanation:

2\cos\left(2x+\dfrac{\pi}{4}\right)=\sqrt2\qquad\text{divide both sides by 2}\\\\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\to2x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+2k\pi\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+2k\pi\ \vee\ 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2k\pi\qquad\text{subtract}\ \dfrac{\pi}{4}\ \text{from all sides}\\\\2x=-\dfrac{2\pi}{4}+2k\pi\ \vee\ 2x=2k\pi\qquad\text{divide all sides by 2}\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=k\pi\ \text{for}\ k\in\mathbb{Z}

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3 years ago
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