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Ipatiy [6.2K]
3 years ago
12

HELP me please!!!!!!!!

Mathematics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer:

I think it is letter C.

eimsori [14]3 years ago
3 0
C. he cannot buy more than 4. He cannot buy a negative amount and he can’t buy a fractional amount.
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pls help with this immediately!!! (only answer if u know bc i cannot get this wrong so don’t try guessing. i’ll be giving braini
arsen [322]

Answer:

your answer should be $16.50 :)

Step-by-step explanation:

5 0
3 years ago
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38 x 97 help help help
Free_Kalibri [48]

Answer:

3686

Step-by-step explanation:

5 0
3 years ago
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Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.
Maru [420]

Answer:

9\text{ln}|x|+2\sqrt{x}+x+C

Step-by-step explanation:

We have been an integral \int \frac{9+\sqrt{x}+x}{x}dx. We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx

\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx

Now, we will apply the sum rule of integrals as:

\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx

9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx

Using common integral \int \frac{1}{x}dx=\text{ln}|x|, we will get:

9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx

Now, we will use power rule of integrals as:

9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx

9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2\sqrt{x}+\int 1dx

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

9\text{ln}|x|+2\sqrt{x}+x+C

Therefore, our required integral would be 9\text{ln}|x|+2\sqrt{x}+x+C.

4 0
3 years ago
Nendell saw the following sign
Phoenix [80]
What is the following sign ??? and the rest of the question I'm sure I can help please provide more info
5 0
3 years ago
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Please help, this is math.
musickatia [10]

Its the second one :)

4 0
3 years ago
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