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Basile [38]
3 years ago
12

(2a2 + ab + 2b) + (4a2 − 3ab + 9)

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
6 0
2a² + ab + 2b + 4a² - 3ab + 9

combine like terms:
2a² + 4a² + ab - 3ab + 2b + 9

6a² - 2ab + 2b + 9
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In the following equation the letter x is known as the:<br> 3x+1=10
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Step-by-step explanation:

First, Take 1 away from 10. 10 - 1 = 9. Then divide 9 by three. 9 divided by three is 3, so X = three.

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Use the given x and y values to write a direct variation<br> equation.<br> x = 5, y = 25
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3 0
3 years ago
100 PTS!! PLEASE HELP! OVER DUE
Reika [66]

Answer:

-\frac{1}{2} \vec {c}

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Look at the component form of each vector.

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If one imagined the line that contained each vector, the line for both would have a slope of 1, because \frac{4}{4}=1=\frac{-2}{-2}

Since they have the same slope they are parallel, but since they are in opposite directions, we often call them "anti-parallel" (simply meaning parallel, but in opposite directions).

If two vectors are parallel, one vector can be multiplied by a scalar to result in the other vector.  This means that there is some number "k", such that k \vec{c} = \vec {d}, or equivalently, kc_x=d_x and kc_y=d_y.

If kc_x=d_x and kc_y=d_y, we just need to substitute known values and solve for k:

kc_x=d_x\\k(4)=(-2)\\k=\dfrac{-2}{4}\\k=-\frac{1}{2}

Double checking that k works for the y-coordinates as well:

kc_y=d_y

(-\frac{1}{2}) (4)    ?    (-2)

-2=-2 \text{ } \checkmark

So, -\frac{1}{2} \vec {c} = \vec {d}

6 0
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