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zhenek [66]
3 years ago
13

Can someone help me with number 12 ?

Mathematics
2 answers:
sveticcg [70]3 years ago
6 0
Steps:
3x - 2(0) = 6
3x = 6
X = 2

Plug 2 into the equation

3(2) - 2y = 6
6 - 2y = 6
6 = 6 + 2y
0 = y

If this is wrong try doing it the opposite way which is:

3(0) - 2y = 6
-2y = 6
Y = -3

3x - 2(-3) = 6
3x + 6 = 6
3x = 0
X = 0
Solnce55 [7]3 years ago
3 0

Answer:

3x= 2y +6

and when we put y = 0

then 3x = 6

x = 6/3

x =2

SECOND one , when we put x = 0

0= 2y +6

-6 = 2y

-6/2= y

-3 = y

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Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
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Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

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Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

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Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

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2 years ago
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