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3 years ago

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I had to write something here so ignore what I am going to say. NSDBHJFGDKTRTEWGSFZFGSDYJFH

2 answers:

Andrej [43]3 years ago

7 0

Answer:

OK TY SIR

Step-by-step explanation:

Vinvika [58]3 years ago

3 0

Thanks hehehehehehe hehshdbd

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Victoria has $200 of her birthday gift money saved at home, and the amount is modeled by the function h(x) = 200. She reads abou

san4es73 [151]

Answer:

h(x) * s(x) = 200(1.05)^(x - 1)

Step-by-step explanation:

Our interest equation is s(x) = (1.05)^(x - 1). This is actually a part of a bigger formula for calculating the amount of money accumulated including interest:

A = P(1 + r)^n, where A is the total, P is the principal amount (initial amount), r is the interest rate, and n is the time

Here, we technically already have the (1 + r)^n part; it's just (1.05)^(x - 1). The principle, though, will actually be the 200 because she starts out at $200.

Thus, to combine these, we simply multiply them together to get:

h(x) * s(x) = 200(1.05)^(x - 1)

4 0

3 years ago

Read 2 more answers

In ΔJKL, the measure of ∠L=90°, LJ = 2.8 feet, and JK = 9.3 feet. Find the measure of ∠J to the nearest tenth of a degree.

Inga [223]

Answer:72.5

Step-by-step explanation:

On delta

8 0

3 years ago

Which transformation of y = f(x) moves the graph 7 units to the left and 3 units down?

Svetlanka [38]

Answer:

b) y = f(x - 7) - 3

Step-by-step explanation:

it adds if its goes to the right and subtracts if it goes to the left

And when it goes down then it will also decrease in number.

that is why b) y = f(x - 7) - 3 is the correct answer to put

8 0

3 years ago

Please help me at this wit steps how did you did it

balu736 [363]

Answer:

5. LCM of 7 and 14: <u> </u><u> </u><em><u>1</u></em><em><u>4</u></em><em><u>. </u></em>

multiples of 7: <u> </u><u> </u><u>7</u><u>,</u><u> </u><u>1</u><u>4</u><u> </u>

multiples of 14: <u> </u><u>1</u><u>4</u><u> </u>

LCM of 8 and 12: <u> </u><u> </u><em><u>2</u></em><em><u>4</u></em><em><u>. </u></em>

multiples of 8: <u> </u><u> </u><u>8</u><u>,</u><u> </u><u>1</u><u>6</u><u>,</u><u> </u><u>2</u><u>4</u><u> </u>

multiples of 12: <u> </u><u> </u><u>1</u><u>2</u><u>,</u><u> </u><u>2</u><u>4</u><u> </u>

Step-by-step explanation:

$.$

5 0

3 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:

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

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

7 0

3 years ago