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2 years ago

PLEASE PLEASE PLEASE HELP

Mathematics

1 answer:

allochka39001 [22]2 years ago

8 0

10- 9

11- 5

12- 1

13- 10

14- 3

15- 7

16- 3

17- 3

18- 6

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Which is f(6) for the quadratic function graphed?<br> 0 -2<br> * -0.5<br> 1.5<br> 4

rusak2 [61]

Answer:

Step-by-step explanation:

I'm pretty sure theres a graph to this so....

The given parabolic graph of quadratic function

To find:What is F(6) for the quadratic function

Solution:We have given graph of quadratic function

there is a different value for Y and a different value for X

For f(6)

here X=6, need to find graph of quadratic function

X=6

f(6)=4

Y=4

Your welcome :)

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3 years ago

A department store buys 300 shirts for a total cost of $7,200 and sells them for $30 each. Find the percent markup.

Novosadov [1.4K]

Answer:

80%

Step-by-step explanation:

because 7200 / 300 equals 24 and 24 / 30 equals .8 so its 80% and if you want to check the answer you take 30 * .8 OR 80% and it gives you 24

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3 years ago

Read 2 more answers

The newspaper says the only 42% of registered voters voted to determine a school bond election. If 11,960 people voted, how many

umka2103 [35]

42% of the body voted = 11960
Total student body = 11960/42% = 28476.19 = 28476 students

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4 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

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3 years ago

A student weighs a sample of the shale and

tia_tia [17]

interesting choice of pfp

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3 years ago