Answer:
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 \\\\\\](https://tex.z-dn.net/?f=%5C%5C%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5E%7Bx%2By%2Bz%7D%7D%20%5C%2C%20dzdydx%20%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5E%7Bx%2By%7De%5Ez%7D%20%5C%2C%20dz%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7D%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5Ez%7D%20%5C%2C%20dz%5Ddydx%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7De%5Ez%5CBig%7C_0%5E%7Bx%2By%7D%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7De%5E%7Bx%2By%7D-e%5E%7Bx%2By%7D%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%2B2y%7D-e%5E%7Bx%2By%7Ddydx%20%5C%5C%5C%5C%5C%5C)
![\\=\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 \\\\](https://tex.z-dn.net/?f=%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%7De%5E%7B2y%7D-e%5E%7Bx%2By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%7De%5E%7B2y%7Ddy-%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7De%5E%7Bx%7De%5E%7By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5Cu%3D2y%5C%5Cdu%3D2dy%5C%5Cdy%3D%5Cfrac%7B1%7D%7B2%7Ddu%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%5Cint%20e%5E%7Bu%7Ddu-%20e%5Ex%5Cint%5Climits%5E%7Bx%7D_%7B0%7De%5E%7By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%5Ccdot%20e%5E%7B2y%7D%5CBig%7C_0%5Ex-%20e%5Exe%5E%7By%7D%5CBig%7C_0%5Ex%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%2B2y%7D%7D%7B2%7D%20-%20e%5E%7Bx%2By%7D%5CBig%7C_0%5Ex%5Ddx%20%5C%5C%5C%5C)
![\\=\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 \\\\\\](https://tex.z-dn.net/?f=%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7D%20-%20e%5E%7B2x%7D-%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%20%2B%20e%5E%7Bx%7D%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7D%20-%5Cfrac%7B3e%5E%7B2x%7D%7D%7B2%7D%20%2B%20e%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7Ddx%20-%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%5Cfrac%7B3e%5E%7B2x%7D%7D%7B2%7Ddx%20%2B%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7De%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5Cu_1%3D4x%5C%5Cdu_1%3D4dx%5C%5Cdx%3D%5Cfrac%7B1%7D%7B4%7Ddu_1%5C%5C%5C%5C%5Cu_2%3D2x%5C%5Cdu_2%3D2dx%5C%5Cdx%3D%5Cfrac%7B1%7D%7B2%7Ddu_2%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20e%5E%7Bu_1%7Ddu_1%20-%5Cfrac%7B3%7D%7B4%7D%5Cint%20e%5E%7Bu_2%7Ddu_2%20%2B%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7De%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5C)
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.
Answer:
The right options are A and D.
Step-by-step explanation:
1. Option A.
The action is:Add 3 to both sides. This means that we need to do the following operation:
x - 3 = 12
∴ x - 3 + 3 = 12 + 3
∴ x = 15
So, the solution of this problem is x = 15.
2. Option D.
The property is:Addition property of equality. This is the property we are chosen because we need to isolate the variable x. To do that, we need to add a number of each side of the equation to find our goal and this property allows us to get our goal.
Answer:
75th term = 443
Step-by-step explanation:
Answer:
.
Step-by-step explanation:
Start by separating this decimal number into its integer part and its fraction part:
The most challenging task here is to express 
as a proper fraction. Once that fraction is found, expressing the original number 
will be as simple as rewriting a mixed number as an improper fraction.
Let 
. 
would then represent the original number.
Note that the repeating digits appear in groups of two. Therefore, if the digits in 
are shifted to the left by two places, the repeating part will continue to match:
.
Note, that this "shifting" is as simple as multiplying the initial number by 
(same as 
raised to the power of the number of digits that needs to be shifted.)
Subtract the original number from the shifted number to eliminate the fraction part completely:
.
In other words:
.
.
Therefore, the original number would be:
.
Answer:
the simple interest in both cases is 200 and 756 respectively
Step-by-step explanation:
The computation of the simple interest is shown below:
As we know that
Simple interest = P × r% × t
So
a. Simple interest is
= 2,500 × 8% × 1
= 200
b. The simple interest is
= 4,200 × 6% × 3
= 756
Hence, the simple interest in both cases is 200 and 756 respectively
