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3 years ago

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:
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3 years ago

Write a fraction that can be used to represent the same value of the decimal number 0.78

avanturin [10]

Answer:

$\frac{78}{100}$

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3 years ago

What is the range of f(x)=4•0.5^x?

VikaD [51]

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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

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

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

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

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

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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

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3 years ago