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MAXImum [283]
3 years ago
14

Help with 8 would be really appreciated!!

Mathematics
2 answers:
Mrac [35]3 years ago
4 0
Answer:
(4,2)
because when x=4, we got
2y=-4+8
y=4/2
=2
klemol [59]3 years ago
3 0
None of these are solutions. because of the points given satisfy the equation.
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8 0
3 years ago
Which graph shows the point (-4, 0)?<br> A. <br> B. <br> C. <br> D.
galina1969 [7]

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5 0
3 years ago
True or False: If circumference of a circle is 12*3.14, then its area is 36*3.14.
strojnjashka [21]

Answer: true

Step-by-step explanation:

8 0
4 years ago
Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0
Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is x^2 - 36 = 0

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

x^{2}-12 x+36=0  --- eqn 1

x^{2}+12 x-36=0 -- eqn 2

x^{2}-36=0  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is x^2 - 36 = 0

3 0
3 years ago
Read 2 more answers
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