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kow [346]
3 years ago
11

Write the equation of the parabola that has the vertex at point (3,−12) and passes through the point (0,6).

Mathematics
2 answers:
Neporo4naja [7]3 years ago
8 0

Answer:

y = 2(x - 3)² - 12

y = (-4/9)(x - 2)² + 7

Step-by-step explanation:

y = a(x - h)² + k

A) y = a(x - 3)² - 12

6 = a(0 - 3)² - 12

18 = 9a

a = 2

y = 2(x - 3)² - 12

B) y = a(x - 2)² + 7

3 = a(-1 - 2)² + 7

-4 = 9a

a = -4/9

y = (-4/9)(x - 2)² + 7

lana [24]3 years ago
3 0

Answer:

y = 2(x-3)^2 -12

y = -4/9(x-2)^2 +7  bonus

Step-by-step explanation:

The vertex form of a parabola is

y = a(x-h)^2 + k  where (h,k) is the vertex

y = a(x-3)^2 - 12

We have one point given (0,6)

6 = a (0-3) ^2 -12

6 = a (-3)^2 -12

6 = 9a-12

Add 12 to each side

6+12 = 9a

18 = 9a

Divide each side by 9

18/9 = 9a/9

a=2

y = 2(x-3)^2 -12

We follow the same steps for the bonus

y = a(x-2)^2 +7

Substitute the point into the equation

3 = a (-1-2)^2 +7

3 =a (-3)^2 +7

3 = 9a +7

subtract 7 from each side

3-7 = 9a +7-7

-4 = 9a

Divide by 9

-4/9 =a

y = -4/9(x-2)^2 +7

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HURRY!!!!!!!!!<br><br> Given P(5,10,8) and Q(7,11,10), find the midpoint of the segment PQ.
rusak2 [61]

Answer:

P_{m}=(6,10.5,9)

Step-by-step explanation:

The mid point can be found with the formula

P_{m}=(\frac{x_{1}+x_{2} }{2},\frac{y_{1} +y_{2} }{2} ,\frac{z_{1}+z_{2}  }{2} )

The given coordinates are P(5,10,8) and Q(7,11,10).

Replacing coordinates in the formula, we have

P_{m}=(\frac{5+7}{2},\frac{10+11 }{2} ,\frac{8+10}{2} )=(\frac{12}{2},\frac{21 }{2} ,\frac{18}{2} )\\P_{m}=(6,10.5,9)

Therefore, the mid point of the segment PQ is P_{m}=(6,10.5,9)

4 0
3 years ago
Question is in the picture
Thepotemich [5.8K]

Answer:

$9.00

Step-by-step explanation:

If the price is at 30% discount, this means it is 100-30 = 70% of the original price.

70 \%  \ of \ original \ price = \$ 6.30\\\\\ original \ price = 6.30 \div 70 \%\\\\= 6.30 \times \frac{100}{70}\\\\=\frac{630}{70}\\\\= 9

Hence, original price is $9.00.

3 0
3 years ago
Given: BC bisects DBE. Prove: ABD is congruent to ABE
Arada [10]

Answer:

See below

Step-by-step explanation:

BC bisects <DBE and if AC is a straight line then you have that:

<ABD + <DBC = 180 (straight angle because of line AC)

<ABE + <CBE = 180 ( straight angle because of line AC)

Because BC bisects <DBE =>   < DBC = <CBE

So <ABD and <ABE must be the same to both sum 180 when added < DBC

7 0
3 years ago
I need to know the answer
makkiz [27]

Step-by-step explanation:

3k + 2k - 6 = 8k + 3

8k - 3k - 2k = - 6 - 3

8k - 5k = -9

3k = - 9

k = - 9/ 3

K = - 3

Hope it will help you :)

3 0
3 years ago
Read 2 more answers
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