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san4es73 [151]
3 years ago
14

Micah is buying items for a birthday party. If he wants to have the same amount each item, what is the least number of packages

of cups he needs to buy?
Mathematics
2 answers:
Simora [160]3 years ago
5 0
That's not enough information. how many of these things are there, do u have a picture of the question?
Alexeev081 [22]3 years ago
5 0

Micah is buying items for a birthday party. If he wants to have the same amount of each item, what is the least number of packages of cups he needs to buy?

A 2 packages. B 3 packages C 5 packages D 4 packages

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The quotient of 38 times a number and -4
nikdorinn [45]

Answer:

140y

Step-by-step explanation:

35 times a number times -4 is 35×y×(-4) which is -140y

4 0
3 years ago
Can someone pls solve this​
denpristay [2]

Answer:

in triangle PQR, <P - <Q = 20 degrees, <Q - <R = 50 degrees, find <P, <Q, <R

Q = 30, R = 80, P = 50

Step-by-step explanation:

in triangle PQR, <P - <Q = 20 degrees, <Q - <R = 50 degrees, find <P, <Q, <R

R  =  Q +50, 80=30+50

P  =  Q + 20, 50=30+20

Q = 30, R = 80, P = 50

3 0
2 years ago
I don’t know the answer I just need some points to ask more questions !!! have a good day !!
Vlad [161]

Answer: Ooh, okay. Have fun then I guess.

7 0
3 years ago
The temperature dropped from 12° to 8° this was drop of how many degrees
lapo4ka [179]

Answer:

4°

Step-by-step explanation:

12-8=4

therefore the answer would be 4 degrees

6 0
3 years ago
Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D
viva [34]

Answer:

\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C

Step-by-step explanation:

So we have the indefinite integral:

\int (\ln(x))^2dx

This is the same thing as:

=\int 1\cdot (\ln(x))^2dx

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}

Simplify:

du=\frac{2\ln(x)}{x}dx

And:

dv=(1)dx\\v=x

Therefore:

\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx

The x cancel:

=x\ln(x)^2-\int2\ln(x)dx

Move the 2 to the front:

=x\ln(x)^2-2\int\ln(x)dx

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

=x\ln(x)^2-2\int 1\cdot\ln(x)dx

Let u be ln(x) and let dv be (1)dx. Thus:

u=\ln(x)\\du=\frac{1}{x}dx

And:

dv=(1)dx\\v=x

So:

=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)

Simplify the integral:

=x\ln(x)^2-2(x\ln(x)-\int (1)dx)

Evaluate:

=x\ln(x)^2-2(x\ln(x)-x)

Now, we just have to simplify :)

Distribute the -2:

=x\ln(x)^2-2x\ln(x)+2x

And if preferred, we can factor out a x:

=x(\ln(x)^2-2\ln(x)+2)

And, of course, don't forget about the constant of integration!

=x(\ln(x)^2-2\ln(x)+2)+C

And we are done :)

8 0
3 years ago
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