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aleksandr82 [10.1K]

3 years ago

14

Help me please above thanks

1 answer:

grigory [225]3 years ago

5 0

Answer:

it might be B.

Step-by-step explanation:

since the answer is 4n i could also be 3n plus a n hopefully is is helpful :3

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I need help,Please it’s due today

Andrej [43]

Answer: Sorry, I’m lazy today so I’ll just answer the first question, letter a. The answer is 30 cups of red paint.

Step-by-step explanation: Since you need 2 cups of blue paint and 4 cups of red paint, the ratio of the amount of cups you need for each paint is 1 cup of blue for every 2 cups of red. You need 15 cups of blue so you multiply it by 2 and you get 30. I hope this wasn’t confusing and I hope this helps. Please listen to you teacher because it would really help you on your work.

7 0

2 years ago

Pedro, Lena, Harriet, and Yermin each plot a point to approximate StartRoot 0.50 EndRoot. Pedro A number line going from 0 to 0.

Scilla [17]

Answer:

Yermin. √0.5 will be at Yermin's point

Step-by-step explanation:

Say Pedro line point is 0.25

Lena is 0.45

Harriet is 0.5

and Yermin is 0.7

The √0.5 = 0.707

≈ 0.7

Therefore it is at Yermin's point

5 0

2 years ago

Read 2 more answers

The equation represents an ellipse.

Tom [10]

Answer:

B) (-10, -6) and (8, -6)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

The larger of two numbers is 1 more than twice the

stellarik [79]

Answer:

Step-by-step explanation:

Hi there!

Let the smaller number be x and the larger number be 2x+1.

According to the question,

The sum of the numbers is 20 less than three

times the larger.

i.e x + (2x+1)= 3*(2x+1) - 20

or, x + 2x +1 = 6x + 3 -20

or, 3x + 1 = 6x - 17

or, 3x = 18

or x= 18/3

x = 6

And 2x+1 = 2*6+1 = 13.

Therefore the smaller number is 6 and larger number is 13.

Hope it helps!

7 0

2 years ago