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3 years ago

Find the roots for f(x) = 2x2 - 8 :

Mathematics

2 answers:

olga_2 [115]3 years ago

7 0

Answer is x=2 or -2 and x=2 or -4

xxTIMURxx [149]3 years ago

5 0

Answer:

x=2 and -2

Step-by-step explanation:

f(x)=2x^2-8

set the function equal to zero to solve for x

2x^2-8=0

add 8 to both sides and divide by 2

x^2=4

take the square root of both sides to get x isolated

√(x^2)=√4

x= ±2

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melamori03 [73]

<u>Answer: </u>

When you buy 2 , 3 and 4 tees , price you pay are 10.33 unit price , 13 unit price and 15.667 unit price respectively.

<u>Solution: </u>

Information provided in the question seems to be less . so assuming following two things.

1)Given function p = n + 5 do not include shipping cost as shipping cost is also dependent on number of tees.

2)If shipping charges for 3 tees is 5 , then shipping charges of 1 tee is $\frac{5}{3}$

When you buy 2 tees , n = 2

Price = p = n+5 = 2+5 = 7

Shipping charges =( $\frac{5}{3}$ ) x 2 = $\frac{10}{3}$

So total price you pay for two tees = 7 + ( $\frac{10}{3}$ ) = 7 + 3.3333 = 10.3333 unit price.

When you buy three tees , n = 3

Price = p = n+5 = 3+5 = 8

Shipping charges =( $\frac{5}{3}$ ) x 3 = 5

So total price you pay for two tees = 8 + 5 = 13 unit price.

When you buy four tees, n = 4

Price = p = n+5 = 4+5 = 9

Shipping charges =( $\frac{5}{3}$ ) x 4 = $\frac{20}{3}$

So total price you pay for two tees = 9 + ( $\frac{20}{3}$ ) = 47/3 = 15.667 unit price.

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4 years ago

Dora goes to a store and buys a phone originally priced at $45. The store is running a discount of 18%. What is the price of the

Alenkinab [10]

Answer: 8.1 $

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3 years ago

Please show all steps... finals coming up!

Rama09 [41]

1) find x
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2) change x on y , and y on x

x= 1/(y+4) - 3
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3 years ago

The y coordinate of the vertex of f (x)=-x^2+8x+7 is f(4)

azamat

Hello there.

To answer this question, we need to remember some properties about the vertex of quadratic functions.

Let $f(x)=ax^2+bx+c,~a\neq0$ . Its vertex can be found on the coordinates $(x_v,~y_v)$ such that $x_v=-\dfrac{b}{2a}$ and $y_v=-\dfrac{b^2}{4a}+c$ , in which $y_v=f(x_v)$ .