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luda_lava [24]
3 years ago
15

Plz help me. how many solutions

Mathematics
2 answers:
Bas_tet [7]3 years ago
8 0

Answer:

no solutions

Step-by-step explanation:

2y = 4x+6

y = 2x+6

Divide the first equation by 2

y = 2x+3

These are parallel lines ( same slope) but different y intercepts

They will never intersect, so they have no solutions

tresset_1 [31]3 years ago
4 0

Answer:

B   No solutions

Step-by-step explanation:

2y = 4x + 6       first equatión

 y  = 2x + 6       second equation

from the first equation

y = (4x+6)/2

y = 4x/2 + 6/2

y = 2x + 3       third equation

matching second equatión and third equation

2x + 6 = 2x + 3

2x - 2x = 3 - 6

0 ≠ -3

then:

Β  No solutions

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Solve the quadratic equation by factoring: 15x^2+4x-4=0
joja [24]
We can factor by grouping. To do so, we multiply the leading coefficient with the constant at the end. In other words, a times c (ax^2 + bx + c).

15*-4 = -60

Now we need to split the b term into two pieces that multiply to -60 and add to 4.

-6 and 10 will work.

Now group one part of b with the 15x^2 and the other part with -4.

(15x^2 + 10x) + (-6x - 4)

Now factor both terms.

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3x+2 is one of our factors and 5x-2 is the other.

(3x+2)(5x-2)=0

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3 years ago
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