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Tpy6a [65]
3 years ago
15

Find the equation of the parabola that has a vertex at (2,0) and a y-intercept of (0,12).

Mathematics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

D

Step-by-step explanation:

standard form would be 3x^2-12x+12

vertex form y=3(x-2)^2

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Mr. Menedez ran 13.1 miles at a constant rate for 150 minutes. To find how fast he ran, you solve the equation 13.1 = x • 150. W
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To find x, divide 13.1 by 150 to get <span>0.08733333333.</span>
5 0
3 years ago
What is the measure of angle Z?
Iteru [2.4K]

Answer:

sum of exterior angles of polygon = 360°

therefore,

z = 360 - (105+145)

z = 360 - 250

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3 0
3 years ago
Let g(x) = 2x2 - 11x. Find g(-1).<br> A) -13 Eliminate <br> B) -9 <br> C) 0 <br> D) 13
worty [1.4K]

Solution

g(x) = 2x^{2} - 11x

Plugging in x = -1 in g(x), we get

g(-1) = 2(-1)^{2} - 11(-1)

Now we have to simplify it.

g(-1) = 2*1 + 11

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g(-1) = 13

So the answer is D) 13.


Thank you. :)

3 0
3 years ago
What number is q in the problem -8-q-10q=-9q+6
uranmaximum [27]
For this question you can say:
-8 -11q = -9q +6 
so now you can add +9q on both sides and +8 on both sides of the equation:
-11q + 9q = 8 + 6 
-2q = 14 
q = 14/-2 
q = -7 :)))
I hope this is helpful
have a nice day
8 0
3 years ago
Read 2 more answers
PLEASE HELP the formula m=12,000+12,000rt/12t gives keri's monthly loan payment where t is the annual interest rate and t is the
tatiyna

Answer:

The answer is below

Step-by-step explanation:

The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.

Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.

m=\frac{12000+12000rt}{12t}\\ but\ m\leq275, \ and \ r=10\%=0.1\\275= \frac{12000+12000(0.1)t}{12t}\\275= \frac{12000}{12t} +\frac{12000(0.1)}{12t}\\275= \frac{1000}{t} + 100\\275-100= \frac{1000}{t} \\175= \frac{1000}{t} \\175t = 1000\\t= \frac{1000}{175}\\ t=5.72\ years

The loan must be at least for 5.72 years for an annual interest rate (r) of 10%

3 0
3 years ago
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