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astra-53 [7]
3 years ago
5

Solve the following quadratic equation for all value of c i'm simplest form 5(x-2)^2=20

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
8 0

Answer:

x=-2 and x=0

Step-by-step explanation:

5(x-2)^2 =  20; divide by 5 both sides

(x-2)^2 = 4; take the square root of both sides

(x-2) = ±2, now set it equal to positive 2 and negative 2

x-2 = 2

x=4

AND

x-2 = -2

x=0

atroni [7]3 years ago
6 0

Answer:

The values of x is 0 and 4.

Step-by-step explanation:

You have to divide 5 by both sides;

5 {(x - 2)}^{2}  = 20

{(x - 2)}^{2}  = 4

You have to expand the brackets;

{x}^{2}  - 4x +4 = 4

{x}^{2}  - 4 x = 0

You have to factorize;

x(x - 4) = 0

Lastly, you have to solve it;

x = 0

x = 4

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3 years ago
What is f(5) if f(1) = 3. 2 and f(x 1) = Five-halves(f(x))?
stealth61 [152]

Function assign value from one set to another. The value of f(5), if f(1) = 3.2 and f(x+1) = Five-halves(f(x)) is 125.

<h3>What is Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

As it is given the value of the function f(1) is 3.2, while the value of f(x+1) is f(x+1) = \dfrac{5}{2}[f(x)], therefore, in order to find the value of f(5), we need to calculate the value of f(4).

f(2)

f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(2)=f(1+1) = \dfrac{5}{2}[f(1)]\\\\f(2) = 2.5 \times 3.2\\\\f(2) = 8

f(3)

f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(3)=f(2+1) = \dfrac{5}{2}[f(2)]\\\\f(3) = 2.5 \times 8\\\\f(3) = 20

f(4)

f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(4)=f(3+1) = \dfrac{5}{2}[f(3)]\\\\f(4) = 2.5 \times 20\\\\f(4) = 50

f(5)

f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(5)=f(4+1) = \dfrac{5}{2}[f(4)]\\\\f(5) = 2.5 \times 50\\\\f(5) = 125

Hence, the value of f(5), if f(1) = 3. 2 and f(x+1) = Five-halves(f(x)) is 125.

Learn more about Function:

brainly.com/question/5245372

8 0
1 year ago
Read 2 more answers
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