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GuDViN [60]
2 years ago
5

Do this jejejejejjejeplsss

Mathematics
1 answer:
In-s [12.5K]2 years ago
4 0

I JDKUHEIINDONT JDJKHS KNOW SKSN

Step-by-step explanation:

JSOSLS

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For 21-22, simplify the algebraic expressions.<br><br> 21. -3(6y) 22. -7a(3b)
Monica [59]

21.

-3(6y)

multiplying we get

-18y

22.

-7a(3b)

on multiplying we get

-21ab

8 0
3 years ago
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Which number line shows the solution to the inequality -2(3x - 1) &lt;8?
Juliette [100K]

Answer:

C

Step-by-step explanation:

I haven't done this math in years but from what I remember this should be the correct answer.

6 0
3 years ago
Write the equation of the line in point slope form that travels through the points (0,5) and (3,7)
Bond [772]

Answer:

y=2/3x+5

Step-by-step explanation:

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4 0
3 years ago
Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

3 0
3 years ago
If the diameter of a circle has endpoints A (7,2) and B (-1,8), where is the center?
docker41 [41]
In the middle: x-coordinate= {7-(-1)}/2= 4 y-coordinate= (8-2)/2= 3 So centre is (4,3)
3 0
3 years ago
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