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.
From Newton’s Second Law of Motion, the force needed to accelerate a moving object of mass m at acceleration a is given by ;
F = ma
In this case, m = 5kg, a = 6m/s^2 and we want to find F
F = 5 x 6
F = 30N
Therefore, a force of 30N is needed to accelerate a 5kg object at 6m/s^2.
Answer:
25 percent
Step-by-step explanation:
7200 divided by 1800 is 25
Answer:
6. x = 28 degrees
7. z = 1.6 cm
Step-by-step explanation:
6.
Notice that you can use the property that tells us that the addition of all internal angles of a triangle must give 180 degrees, then you write the following equation:
50 + 69 + (2 x +5) = 180
combine like terms:
119 + 2 x + 5 = 180
124 + 2 x = 180
subtract 124 from both sides:
2 x = 56
divide by 2 both sides:
x = 56 / 2
x = 28 degrees
Problem 7.
If the two triangles are congruent, then the side MN must equal side RS.
Since MN measure 1,8 cm, then RS must also measure 1.8 cm
and we can write the equation:
1.8 = 3 z - 3
adding 3 to both sides:
1.8 + 3 = 3 z
4.8 = 3 z
dividing both sides by 3:
z = 4.8 / 3
z = 1.6 cm
All you have to do is divde hope this helps thank ylu
