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3 years ago

Find one value of x that is a solution to the equation:(x^2– 8)^2 + x^2 – 8 = 20x=

Mathematics

1 answer:

Kipish [7]3 years ago

8 0

Answer:

x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)

Step-by-step explanation:

Solve for x:

-8 + x^2 + (x^2 - 8)^2 = 20

Expand out terms of the left hand side:

x^4 - 15 x^2 + 56 = 20

Subtract 20 from both sides:

x^4 - 15 x^2 + 36 = 0

Substitute y = x^2:

y^2 - 15 y + 36 = 0

The left hand side factors into a product with two terms:

(y - 12) (y - 3) = 0

Split into two equations:

y - 12 = 0 or y - 3 = 0

Add 12 to both sides:

y = 12 or y - 3 = 0

Substitute back for y = x^2:

x^2 = 12 or y - 3 = 0

Take the square root of both sides:

x = 2 sqrt(3) or x = -2 sqrt(3) or y - 3 = 0

Add 3 to both sides:

x = 2 sqrt(3) or x = -2 sqrt(3) or y = 3

Substitute back for y = x^2:

x = 2 sqrt(3) or x = -2 sqrt(3) or x^2 = 3

Take the square root of both sides:

Answer: x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)

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2 years ago

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PLEASE HELP ASAP.

alexandr1967 [171]

Answer:

$Y' = (-2, 4\frac{2}{3})$

Step-by-step explanation:

Given

$Y = (-3,7)$

$Scale\ Factor = (\frac{2}{3}x,\frac{2}{3}y)$

Required

Determine Y'

Y' can be solved by multiplying the scale factor by Y

i.e.

$Y' = Scale\ Factor * Y$

For, the x coordinates.

$Y' = \frac{2}{3}x$

Where

$x = -3$

$Y' = \frac{2}{3} * -3$

$Y' = \frac{-6}{3}$

$Y = -2$

For the y coordinates:

$Y' = \frac{2}{3}y$

Where

$y = 7$

$Y' = \frac{2}{3} * 7$

$Y' = \frac{14}{3}$

$Y' = 4\frac{2}{3}$

Hence:

$Y' = (-2, 4\frac{2}{3})$

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3 years ago

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Anni [7]

Answer: $\frac{1}{10}$

Step-by-step explanation:

First, we need to find the common denominator

The easiest way to do this is by multiplying the two given denominators, which are 2 and 5.

2 × 5 = 10

So, our common denominator is 10. Then, multiply the numerators of the two fractions by 2 and 5. Here's why:

$\frac{3*2}{5*2}$ = $\frac{6}{10}$ We multiplied the top and bottom by 2 to make sure our new fraction stays equivalent to the original fraction.

Do the same thing for the other one:

$\frac{1*5}{2*5}$ = $\frac{5}{10}$

Finally, subtract the two fractions to find the difference between the two times:

$\frac{6}{10} -\frac{5}{10}$ = $\frac{1}{10}$

The reason we used the common denominator is because we can only add or subtract fractions if they have the same denominator.

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3 years ago

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Select "Growth" or "Decay" to classify each function.

allochka39001 [22]

1. The growth rate equation has a general form of:

y = A (r)^t

The function is growth when r≥1, and it is a decay when r<1. Therefore:

y=200(0.5)^2t --> Decay

y=1/2(2.5)^t/6 --> Growth

y=(0.65)^t/4 --> Decay

2. We rewrite the given equation (1/3)^d−5 = 81

Take the log of both sides:

(d – 5) log(1/3) = log 81

d – 5 = log 81 / log(1/3)

d – 5 = - 4

Multiply both sides by negative 1:

- d + 5 = 4

So the answer is D

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3 years ago

Find one value of x that is a solution to the equation:(x^2– 8)^2 + x^2 – 8 = 20x=​

Find one value of x that is a solution to the equation:(x^2– 8)^2 + x^2 – 8 = 20x=