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Ipatiy [6.2K]
3 years ago
13

I need help on this one please

Mathematics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

it is D

Step-by-step explanation:

just reflect by y=-1 its easy

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If you know how many CD's Mickey orders,can you determine how much money he spends?Write the correspondering expression.
Setler [38]

no not until u know the price of a cd u cant determine the total price
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4 years ago
What are the zeros of the quadratic function f(x) = 2x2 + 16x – 9?
Lena [83]
To find the zeros of this equation you need to first set it equal to zero
2x2 + 16x - 9 = 0
But since you can't FOIL this equation you need to move the non-variable number over
2x2 +16x = 9
Now solve for x by pulling an x out of the equation
x(2x + 16) = 9
x = 9
2x +16 = 9
2x = -7
x = -7/2
So your zeros would be at x = 9, and -7/2
7 0
3 years ago
A nursery has $50,000 of inventory in dogwood trees and red maple trees. The profit on a dogwood tree is 27% and the profit on a
nasty-shy [4]

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Step-by-step explanation:

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2 years ago
How can you quickly calculate 20 percent of any bill amount?
il63 [147K]

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8 0
3 years ago
Read 2 more answers
If a triangle has side lengths 7, 10, and 12, is it a right triangle??
Vesnalui [34]

Answer:

No. Remember, a right angle must have a 90 degree angle. We can find the lengths with the Pythagorean Theorem.

Step-by-step explanation:

Given the length 7, 10, and 12, we can assume that 12 is the hypotenuse (it is the longest length).

- we can use 7 and 10 interchangeably.

Fill in the equation, a^2 + b^2 = c^2

                 where c = 12, and a or b = 7 or 10.

To indicate if the given lengths would form a right angle, we can only input 7 or 10, not both.

Therefore, 7^2 + b^2 = 12^2 or 10^2 + b^2 = 12^2

7^2 + b^2 = 12^2 ==> 49 + b^2 = 144 ==> <u>b= </u>\sqrt{95<u> ==> </u><u>9.746</u>

b= 9.7, not 10.

10^2 + b^2 = 12^2 ==> 100 + b^2 = 144 ==> <u>b = </u>\sqrt{44<u> ==> </u><u>6.633 </u>

b= 6.6, not 7.

Therefore, the lengths 7, 10, and 12, does NOT make a right triangle.

Hope this helps!

5 0
3 years ago
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