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2 years ago

Length=9in width=2in height=3in. what is the surface area

Mathematics

1 answer:

baherus [9]2 years ago

4 0

Answer:

102in2

Step-by-step explanation:

The formula for surface area is 2(wl+hl+hw). So plug in the measurements of the prism into the formula to get 102.

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What grades are considered good?

Dahasolnce [82]

Answer:

Depends who you are and what your standard grades are...

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

3. Andrea is buying snacks at the basketball game. The snacks are $2.25 apiece. She wants to buy 4

soldi70 [24.7K]

Answer:

D - No, the ordered pair (4,9) is a solution to this problem.

Step-by-step explanation:

If you plotted the number of snacks as the x-axis, and then the total cost of the snacks on the y-axis, you would be able to graph a line and see the cost of snacks based on the number purchased. If you were going to write an equation for this, it would be y=2.25x (or y = 2 1/4 x)

In this case, as each snack is the same price, the y-values for each x would be as such: (1, 2.25) <- one snack, $2.25;

(2, 4.5)

(3, 6.75)

(4, 9)

So that is why the ordered pair of (4,9) would be a solution to this problem.

Hope that makes sense!

4 0

3 years ago

13. Write

Bingel [31]

First off, we factor out the expression:

$\displaystyle \large{y = 2 {x}^{2} - 12x + 16} \\ \displaystyle \large{y = 2 ( {x}^{2} - 6x + 8) }$

In the bracket, separate 8 out of the expression.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 8)] }\\ \displaystyle \large{y = 2[ ( {x}^{2} - 6x) + 8]}$

In x^2-6x, find the third term that can make up or convert it to a perfect square form. The third term is 9 because:

$\displaystyle \large{ {(x - 3)}^{2} = {x}^{2} - 6x + 9}$

So we add +9 in x^2-6x.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 9) + 8]}$

Convert the expression in the small bracket to perfect square.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8]}$

Since we add +9 in the small bracket, we have to subtract 8 with 9 as well.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8 - 9]} \\ \displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}$

Then we distribute 2 in.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\$

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\ \displaystyle \large{y = [2 \times {(x - 3)}^{2} ]+[ 2 \times ( - 1)] } \\ \displaystyle \large{y = 2 {(x - 3)}^{2} - 2 }$