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ExtremeBDS [4]
3 years ago
12

Answer this as soon as possible

Mathematics
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

A. -6 - 15 = - 21

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Can someone help me with this? Last one is -1/4 btw
Leya [2.2K]
2 because rise/run so 8/4 which=2
5 0
2 years ago
HELP!!<br>Find the third term of (x^2+3y)^3
lana66690 [7]

Answer:

T_{3}=27{x}^{2}y^2

Step-by-step explanation:

The given binomial expression is:

( {x}^{2} + 3y)^{3}

When we compare to:

{(a +b)}^{n}

We have

a =  {x}^{2}

b = 3y \\ n = 3

The nth term is given by;

T_{r+1}=^nC_ra^{n-r}b^r

To find the 3rd term, we put:

r + 1 = 3 \\ r = 2

We substitute into the formula to get:

T_{3}=^3C_2( {x}^{2} )^{3-2}(3y)^2

We simply:

T_{3}=3( {x}^{2} )^{1} \times 9y^2

T_{3}=27{x}^{2}y^2

3 0
3 years ago
ℜ 0, -180 is the same rotation as ℜ 0, 180. True False
neonofarm [45]

Answer:

Step-by-step explanation:

The first rotation is clockwise, whereas the second is counterclockwise.  We end up at the same point either way, and the trig functions are the same.  

In summary, the final outcomes are the same, but the directions of rotation are different.   For this reason:  False

8 0
3 years ago
✓<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B19%20%2B%20%20%5Csqrt%7B30%20%2B%20%20%5Csqrt%7B32%20%2B%20x%20%20%7D%20%7D%
Setler79 [48]

We'll have to repeatedly square both sides of the equation, in order to get rid of the square roots. Squaring a first time yields

19+\sqrt{30+\sqrt{32+x}}=25

Move the 19 to the right hand side:

\sqrt{30+\sqrt{32+x}}=6

And square again:

30+\sqrt{32+x}=36 \iff \sqrt{32+x}=6

Square one last time:

32+x=36 \iff x=36-32=4

Let's check the solutions: all these squaring might have created external solutions:

\sqrt{19+\sqrt{30+\sqrt{32+4}}}=\sqrt{19+\sqrt{30+6}}=\sqrt{19+6}=\sqrt{25}=5

So, x=4 is a feasible solution.

8 0
3 years ago
If we list all the possible factors (in order) of 12, we get 1, 2, 3, 4, 6, 12. This is the case because 1 x 12, 2 x 6, and 3 x
____ [38]

Answer:

1, 2, 3, 4, 6, 9, 12, 18, and 36, - 1, - 2

7 0
3 years ago
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