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Pachacha [2.7K]
3 years ago
15

I do not understand how to do this problem. I am very confused and need help. x-3/2x+4=-1/2

Mathematics
1 answer:
atroni [7]3 years ago
4 0
X - \frac{3}{2} x + 4 = - \frac{1}{2}

First, simplify \frac{3}{2} x to \frac{3x}{2} / Your problem should look like: x - \frac{3x}{2} + 4 = - \frac{1}{2}
Second, simplify x - \frac{3x}{2} + 4 to - \frac{x}{2} + 4 / Your problem should look like: -\frac{x}{2} + 4 = -\frac{1}{2}
Third, regroup terms. / Your problem should look like: 4 - \frac{x}{2} = - \frac{1}{2}
Fourth, multiply both sides by 2. / Your problem should look like: 8 - x = -1
Fifth, subtract 8 from both sides. / Your problem should look like: -x = -1 - 8
Sixth, simplify -1 - 8 to -9. / Your problem should look like: -x = -9
Seventh, multiply both sides by -1. / Your problem should look like: x = 9

Answer: x = 9

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Answer:

Standard Form           Equivalent Form            Extreme Values

y=x^2-6x+17                   (x-3)^2+8                        (3,8)

y=x^2+8x+21                  (x+4)^2+5                        (-4,5)

y=x^2-16x+60                 (x-8)^2-4                         (8,-4)

Step-by-step explanation:

1) Standard form:

y=x^2-6x+17

Equivalent Form:

Can be found using completing the square method.

y=x^2-6x+17\\y=x^2-2(x)(3)+(3)^2-(3)^2+17\\y=(x-3)^2-9+17\\y=(x-3)^2+8

So, Equivalent form is: (x-3)^2+8

Extreme value:

Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate is: 2x-6

Now, put the derivate equal to zero: 2x-6 = 0

2x=6\\x=6/3 \\x=3

Maximum value can be found by putting minimum value in the given function:

Put x = 3 and solve:

(3)^2-6(3)+17\\9-18+17\\9-1\\=8\\

So, the extreme values is: (3,8)

2) Standard form:

y=x^2+8x+21

Equivalent Form:

Can be found using completing the square method.

y=x^2+8x+21\\y=x^2+2(x)(4)+(4)^2-(4)^2+21\\y=(x+4)^2-16+21\\y=(x+4)^2+5

So, Equivalent form is: (x+4)^2+5

Extreme value:

Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate of x^2+8x+21 is: 2x+8

Now, put the derivate equal to zero:

2x+8 = 0\\2x=-8\\x=-8/2 \\x=-4

So, minimum value is: -4

Maximum value can be found by putting minimum value in the given function:

Put x = -4 and solve:

x^2+8x+21\\=(-4)^2+8(-4)+21\\=16-32+21\\=5

So, Maximum value is: 5

So, the extreme values is: (-4,5)

3) Standard form:

y=x^2-16x+60

Equivalent Form:

Can be found using completing the square method.

y=x^2-16x+60\\y=x^2-2(x)(8)+(8)^2-(8)^2+60\\y=(x-8)^2-64+60\\y=(x-8)^2-4

So, Equivalent form is: (x-8)^2-4

Extreme value:

Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate of x^2-16x+60 is: 2x-16

Now, put the derivate equal to zero:

2x-16 = 0\\2x=16\\x=16/2 \\x=8

So, minimum value is: 8

Maximum value can be found by putting minimum value in the given function:

Put x = 8 and solve:

x^2-16x+60\\=(8)^2-16(8)+60\\=64-128+60\\=-4

So, Maximum value is: -4

So, the extreme values is: (8,-4)

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