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Alex_Xolod [135]
3 years ago
5

What’s the answer to this problem that I can not figure out

Mathematics
1 answer:
Maslowich3 years ago
4 0

Answer:

-64/3 HOPE THIS HELPS if you need more help let me know

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20 is 1/10 (one tenth) of what number?
Tom [10]

Answer:

200

Step-by-step explanation:

200/10=20

6 0
3 years ago
Someone please help me :((
Dmitrij [34]
8x^2-3 = sqrt(16x+9)
[8x^2-3]^2 = [sqrt(16x+9)]^2 ... square both sides
64x^4-48x^2+9 = 16x+9
64x^4-48x^2+9-16x-9 = 0
64x^4-48x^2-16x = 0
16x(4x^3-3x-1) = 0
16x(x-1)(2x+1)^2 = 0
16x=0 or x-1=0 or (2x+1)^2 = 0
x=0 or x=1 or x = -1/2

The possible solutions are x=0 or x=1 or x = -1/2

We need to check all the possible solutions

Checking x=0
8x^2-3 = sqrt(16x+9)
8(0)^2-3 = sqrt(16*0+9)
-3 = 3
The equation is false so x=0 is extraneous (not a real solution)

Checking x=1
8x^2-3 = sqrt(16x+9)
8(1)^2-3 = sqrt(16*1+9)
5 = 5
Equation is true. The value x=1 is a solution

Checking x=-1/2
8x^2-3 = sqrt(16x+9)
8(-1/2)^2-3 = sqrt(16(-1/2)+9)
-1 = 1
The equation is false so x=-1/2 is extraneous (not a real solution)

Therefore, the only answer is choice A) 1
6 0
4 years ago
ANSWER QUICK GIVING BRAINIEST!
Alenkinab [10]

Answer: yes because you wouldn’t be able to multiply them and they wouldn’t bothe be fractions.

Step-by-step explanation:

8 0
3 years ago
The perimeter of a rectangle is 400 yards what are the dimensions of the rectangle if the length is 40 yards more than the with
zloy xaker [14]
P = 2(L + W)
P = 400
L = W + 40

400 = 2(W + 40 + W)
400 = 2(2W + 40)
400 = 4W + 80
400 - 80 = 4W
320 = 4W
320/4 = W
80 = W.....so the width (W) = 80 yards

L = W + 40
L = 80 + 40
L = 120...and the length (L) = 120 yards
7 0
3 years ago
Determine whether 606 is divisible by 2, 3, 4, or 6.
Serggg [28]
606/2=303 \\ 
 \\ 606/3=202 \\ 
 \\ 606/4=151.5 \\ 
 \\ 606/6=101 \\ 
 \\ Answer:C.2,3 \ and \ 6 \ only \\
4 0
4 years ago
Read 2 more answers
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