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VLD [36.1K]
3 years ago
7

In the parallelogram drawn for the parallelogram method, what does the diagonal between the two terminal points of the vectors r

epresent? Explain your answer.
Mathematics
1 answer:
DIA [1.3K]3 years ago
7 0
<span>The difference of the vectors; sample: the other diagonal would be the sum of one of the vectors and the opposite of the other vector, so it would be the difference.

</span>
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Octagon A has a side length of 4 inches, and an area of 32 square inches. Octagon B is similar to octagon A, and has a correspon
Anika [276]

Answer:

  128 square inches

Step-by-step explanation:

The scale factor for area is the square of the scale factor for side length. The larger octagon has an area of ...

  (8/4)²(32 in²) = 4·32 in² = 128 in²

The area of octagon B is 128 square inches.

6 0
3 years ago
Find the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B25%20%5Ctimes%202%7D%20%20%5Ctimes%20%2
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5 0
3 years ago
Which expression shows the result of applying the distributive property to 3(2x−6) ?
Serga [27]
6x-18

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8 0
4 years ago
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In gym class, each student runs 7 laps. There are 8 students in each gym class. How many laps will be run by one class during 2
Neko [114]

Answer: 112


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3 0
4 years ago
I WILL GUVE BRAINLIEST PLEASE HELPPPP pLz algebra one
olga55 [171]

Answer:

\displaystyle L=4x^{4}y^{8}z^{12}

Step-by-step explanation:

<u>The Volume of Rectangular Prism</u>

Given a rectangular prism of width W, height H, and length L, its volume is calculated as follows:

V = WHL

We are given the volume of a rectangular prism:

V=72x^8y^{14}z^{22}

It's also known the width is:

W=6x^3y

And height

H=3xy^5z^{10}

Substituting in the formula, we solve for L:

72x^8y^{14}z^{22}=(6x^3y)*(3xy^5z^{10})L

\displaystyle L=\frac{72x^8y^{14}z^{22}}{(6x^3y)*(3xy^5z^{10})}

Operating the denominator

\displaystyle L=\frac{78x^8y^{14}z^{22}}{18x^4y^6z^{10}}

Dividing:

\displaystyle L=4x^{8-4}y^{14-6}z^{22-10}

Simplifying:

\boxed{\displaystyle L=4x^{4}y^{8}z^{12}}

6 0
3 years ago
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