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4 years ago

What is -15+v over -3=21

Mathematics

1 answer:

Snezhnost [94]4 years ago

4 0

Answer:

V = -48

Step-by-step explanation:

((-15 + v) / -3) = 21

(-15 + v) = -63

v = -63 + 15

v = -48

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5^-4 over 5^3 simplify A. 5^7 B. 5^-1 C. 1/5 D. 1/5^7

GuDViN [60]

Answer:

D is the correct answer.

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3 years ago

Erin is considering joining one of 2 clubs. Club A has no registration fee, but charges $105 per month. Club B charges members $

inn [45]

Answer:

4 months

Step-by-step explanation:

Club A

registration fee: $0

monthly fee: $105

After every month, the total cost increases by $105.

month 0: $0

month 1: $105

month 2: $210

month 3: $315

month 4: $420

month 5: $525

month 6: $630

Club B

registration fee: $375

monthly fee: $80

Notice how Club B's total reaches Club A's total after 2 months.

month 0: $375

month 1: $455

month 2: $535

month 3: $615

month 4: $695

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4 years ago

Given that y = sin(x+y),find the derivative when (x,y)=(π,0)

lisov135 [29]

<h2>Answer:</h2>

Shown in the explanation

<h2>Step-by-step explanation:</h2>

Recall that an implicit function is a relation given by the form:

$R(x_{1},\ldots, x_{n})=0$

Where $R$ is a function of two or more variables. In this case, that function is:

$y = sin(x+y)$

and is implicit because we can define it as:

$y-sin(x+y)=0$ having two variables.

So, let's take the derivative:

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right) \\ \\$

Applying chain rule:

$\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)$

But:

$\frac{d}{dx}\left(y\right)=y'$

Therefore:

$y'=\cos \left(x+y\right)\left(1+y'\right)$

Isolating $y'$ :

$\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}$

When $(x,y)=(\pi,0)$ :

$\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}$