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OLga [1]
3 years ago
12

PLEASE HELP 2x^2-8x+6=0

Mathematics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

x = 4 or x = -1

Step-by-step explanation:

2x^2 - 8x + 6 =0

divide by 2 we get  : x^2 -4x +3 = 0

(x -4) (x+1)

x = 4 or x = -1

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1,290=h/10+h/5 What is h?​
alekssr [168]

Answer:

4300

h/10 + 2h/10

= 1290

3h/10 = 1290

3h = 12,900

h = 4300

6 0
3 years ago
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Uhh so i need help with this , it would be very nice if ya’ll helped a sis out
Masja [62]

9514 1404 393

Answer:

  (c)  (3, 6)

Step-by-step explanation:

The only point that is on the line is (3, 6).

__

I find a graphing calculator to be a useful tool.

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For solving this "by hand," you put the x- and y-values into the equation and check to see if it is true.

A. 2(0) +3(12) = 36 ≠ 24

B. 2(2) +3(9) = 31 ≠ 24

C. 2(3) +3(6) = 24 . . . . . a solution

D. 2(8) +3(0) = 16 ≠ 24

3 0
3 years ago
Please help? I'm. terrible at these
kykrilka [37]
13.5 = 30% of 45
4.5 = 15% of 30
4.6 = 23% of 20
4.2 = 60% of 7

Hope this helps!!
3 0
3 years ago
Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i
lana66690 [7]

In the x,y plane, we have z=0 everywhere. So in the equation of the sphere, we have

25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2

which is a circle centered at (2, -10, 0) of radius 4.

In the x,z plane, we have y=0, which gives

25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the y,z plane, x=0, so

25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2

which is a circle centered at (0, -10, 3) of radius \sqrt{21}.

3 0
3 years ago
5d-4=4-d<br> solve the equation
horrorfan [7]

Answer:   d = 0


Step-by-step explanation:

5d - 4 = 4 - d

+d             +d

____________

6d - 4 = 4

      -4   -4

__________

6d = 0

__   __

6      6

d = 0

3 0
3 years ago
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