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Svet_ta [14]
3 years ago
13

I dont know what the asymptote is so i cant write the equation​

Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

here is the answer to the question

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5. If Amber decides to mail copies of her photos to the 45 residents of her grandmother’s assisted living facility, the new func
pav-90 [236]

Answer: please find the answer in the explanation

Step-by-step explanation:

From the information given from the question, the initial function representing her photo sharing is:

F(x) = 3(4)^2

F(x) = 3 × 16 = 48 photos

Then the new graph has an upward shift ( translation) of 45 units

The new function representing her photo sharing is:

F(x) = 3(4)^2 + 45

F(x) = 48 + 45

F(x) = 93 photos.

8 0
3 years ago
Read 2 more answers
Four pounds of chicken is divided into 1/3 -pound packages. How many packages of chicken are there?
Katen [24]

Answer:12

Step-by-step explanation:

there is 4 pounds total, and each package is 1/3 a pound, so 3 packages, (1/3x3)=3 so 3 x 4 for all packges= 12. Hope this helps!

4 0
4 years ago
What is the length, in units, of the line segment with endpoints at (1, 4) and (3, 7)?
uranmaximum [27]

Answer:

√13

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Please see attachment
amm1812

Answer: \frac{2(-\pi^5) }{5} + 18 + C , I hope this is right. Im kind of new to integral calculus

Step-by-step explanation:

You essentially want to split this integral into two parts, because its a piecewise function. The conditions on the right side basically say for when x is less than and when x is greater than 0. So, we should split this integral at 0.

The first function says that -pi is less than 0, so change the upper boundary to 0. This make the definite integral bounded at -pi to 0. But now, we have to take the anti derivative of 2x^4 using the reverse power rule. Add one to the power, then divide by the new power. So the new function is \frac{2x^5}{5} defined at lower boundary -pi and upper boundary of 0. Use the fundamental theorem of calculus to solve the integral. You get \frac{2(-\pi ^5)}{5} - \frac{2(0^5)}{5} = \frac{2(-\pi^5) }{5}. You want to add the 2nd integral to this, so lets go solve that.

Now for this integral, we have the conditions that pi is greater than 0. So 0 will be in the lower boundary this time and pi will be in the upper. Basically apply the same thing from the first integral. Do anti derivative and use fund. theorem of calculus. the anti-derivative of sinx is -cosx. You can take this 9 out because its a constant that is scaling -cosx. So now 9 is on the outside of the integral and you have the definite integral of -cosx with the boundaries of 0 and pi. Plug in pi to get -1, then multiply that by the negative sign to get postive 1. Now let's plug in the lower boundary, 0. Cosine of 0 is 1, then we are multiplying by the negative sign to get -1. Subtract 1 and -1 to get postive 2. Finally multiply that by the 9 from before to get 18

Final Answer finally:

\frac{2(-\pi^5) }{5} + 18

OH, do not forget the C. This took me 30 minutes of constant typing so hopefully Im right

7 0
3 years ago
Find the missing side lengths. Leave your answers as radicals in simplest form.
ryzh [129]

Answer:

D.

u =3

v =\sqrt{3}

Step-by-step explanation:

Given

The triangle in the diagram above

Required

Find the missing lengths, u and v

To find the missing lengths, we have to check the relationship between the missing lengths, the given length and the given angle;

From trigonometry;

sin\ \theta = \frac{Opp}{Hyp}; Where Opp = Opposite and Hyp = Hypotenuse

From the attached diagram;

\theta = 30

Opp = v

Hyp = 2\sqrt{3}

sin\ \theta = \frac{Opp}{Hyp}  becomes

sin\ 30 = \frac{v}{2\sqrt{3}}

Multiply both sides by 2\sqrt{3}

2\sqrt{3} * sin\ 30 = \frac{v}{2\sqrt{3}} * 2\sqrt{3}

2\sqrt{3} * sin\ 30 = v

In radians, sin30 = \frac{1}{2}

2\sqrt{3} * \frac{1}{2} = v

\sqrt{3} = v

v =\sqrt{3}

Similarly, From trigonometry;

cos\ \theta = \frac{Adj}{Hyp}; Where Adj = Adjacent

From the attached diagram;

\theta = 30

Adj = u

Hyp = 2\sqrt{3}

cos\ \theta = \frac{Adj}{Hyp}  becomes

cos\ 30 = \frac{u}{2\sqrt{3}}

Multiply both sides by 2\sqrt{3}

2\sqrt{3} * cos\ 30 = \frac{u}{2\sqrt{3}} * 2\sqrt{3}

2\sqrt{3} * cos\ 30 = u

In radians, cos30 = \frac{\sqrt{3}}{2}

2\sqrt{3} * \frac{\sqrt{3}}{2} = u

\sqrt{3} * \sqrt{3} = u

3 = u

u =3

3 0
3 years ago
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