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Tcecarenko [31]
3 years ago
5

Sarah went on a run. On her run she found 6 rocks. When she was going back home she found 7 more rocks. When she got home she re

lized she lost 2 of the rocks. How many rocks does she have now?
Mathematics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

11

Step-by-step explanation:

first you add 6 + 7 = 13

then you 13 - 2 which gives you eleven!

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Evaluate triple integral ​
kaheart [24]

Answer:

\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}

Step-by-step explanation:

\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\

\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\

\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\

\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

6 0
3 years ago
How to make 9/7 to a simplified fraction?
Leviafan [203]

Divide 7 by 9 and you’ll get your answer in a calculator

8 0
3 years ago
If l || m what is the value of y??
allsm [11]

Answer:

b

Step-by-step explanation:

7 0
3 years ago
I need help!!!! Please lol
Dmitry_Shevchenko [17]
I hope this helps you

4 0
3 years ago
Determine which equation is belongs to the graph of the limacon curve below.
JulsSmile [24]

Answer: Option C.

Step-by-step explanation:

We have functions of r(θ)

In our graph, we can see that the minimum value of r is when θ = 0°, and the maximum value is when θ = 180°.

We also know that the graph is in the square (-5, 5)x(-5, 5) and you can see that the maximum radius is almost less than 5.

Then let's analyze the options:

A) r = 3 - 2*cos(θ)

the maximum is at the right angle, but the maximum is:

r = 3 -2*(-1) = 5, so this maximum value is bigger than the one in the graph.

B) r = 3 - sin(θ)

For the sin functions, the maximum and minimum do not correspond with the values i write earlier, so we can discard this option.

C) r = 3 - cos(θ)

The maximum is: r = 3 - (-1) = 4, so this may be the correct answer.

the minimum is r = 3 - 1 = 2,

this is a possible equation for our circle.

D) r = 2 - 2*cos(θ)

Here, when θ = 0, we have: r = 2 - 2*1 = 0, but in the graph we can see that the radius is not 0 when θ = 0, so we can discard this option.

So the only option that makes sense is option C.

8 0
3 years ago
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