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Marina CMI [18]
3 years ago
7

Looking for the best answer on this

Mathematics
1 answer:
Lana71 [14]3 years ago
4 0

Answer:

\tt B)\: \cfrac{x+3}{x-3}

Step-by-step explanation:

\tt \cfrac{5x+15}{5x-15}

Factor out the common term 5

5x+15 → 5(x+3)

5x-15 → 5(x-3):-

\tt \cfrac{5\left(x+3\right)}{5\left(x-3\right)}

Cancel the common factor → 5

\tt \cfrac{x+3}{x-3}

Therefore, your answer is B!!

_____________________

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Seth is reading a long novel. if he reads 45 pages per hour, how many hours will it take him to read 585 pages?
vitfil [10]
It will take Seth 13 hours to read the novel.
This is true because:
You need to divide 585 by 45, and you get the answer, 13.
6 0
3 years ago
Write a fraction that can be used to represent the same value of the decimal number 0.78
avanturin [10]

Answer:

\frac{78}{100}

If you need a simplified fraction then \frac{39}{50}


8 0
3 years ago
What is the range of f(x)=4•0.5^x?
VikaD [51]

The range is the output, which would be the Y value.

The Y value would be calculated based on the input value for x.


any positive number raised to any value of x would be a positive number so 0.5^x would always be positive.

That would be multiplied by 4, which is a positive value, so the output ( Y value) would always be positive.


The range for this would be (0,∞)

6 0
3 years ago
Why is the answer to this integral's denominator have 1+pi^2
ss7ja [257]

It comes from integrating by parts twice. Let

I = \displaystyle \int e^n \sin(\pi n) \, dn

Recall the IBP formula,

\displaystyle \int u \, dv = uv - \int v \, du

Let

u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Then

\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn

Apply IBP once more, with

u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Notice that the ∫ v du term contains the original integral, so that

\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)

\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I

\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n

\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C

6 0
2 years ago
This diagram of a rectangular city park was drawn using a scale of 1 centimeter to 20 meters
I am Lyosha [343]

Answer:

Scale factor for the drawing from actual park = \frac{1}{20}

Area of the actual park = 1200 cm²

Step-by-step explanation:

Length of the rectangular city park = 5 cm

Width of the rectangular park = 6 cm

Using scale factor 1 cm = 20 meters

Scale factor = \frac{\text{length of the park in drawing}}{\text{Actual length of park}}=\frac{1}{20}

\frac{5}{\text{Actual length of park}}=\frac{1}{20}

Actual length = 5 × 20 = 100 meters

Actual width = 6 × 20 = 120 meters

Area of the rectangular park = length × width

                                               = 100 × 120

                                               = 1200 square meters

Therefore, Scale factor from actual length to the length in drawing = 1 : 20

Area of the rectangular park = 1200 square feet

7 0
3 years ago
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