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3 years ago

How is your day? also for the sake of asking, whats 5×1Love yallllll!

Mathematics

2 answers:

san4es73 [151]3 years ago

7 0

My day was good :>

uh it’s 5 does your brain need a break?

:>

MA_775_DIABLO [31]3 years ago

3 0

Answer:

My day is great, how is yours?

Step-by-step explanation:

5x1 is 5.

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What is the distance of KF. round to the nearest tenths

snow_tiger [21]

Answer:

its 9.2

Step-by-step explanation:

Well i counted 7 units (x-axis) and 6 units (y axis). 6 squared is 36 and 7 squared is 49. Add them together and get 85. Square root 85 and youll get 9.21954445729. Round it to the nearest tenths and get 9.2

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3 years ago

Guided Practice #3:

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3 years ago

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

Use the equation of the water level of the river represented by the equation y=-4x + 170, where x represents the

vovangra [49]

Answer:

(0,170) and (60,-70) and (5, 150)

Step-by-step explanation:

To see if a given point is on the line or not, you just have to enter the x value (first value in the parenthesis) and see if the function returns the correct y value (second number in the parenthesis).

f(x) = -4x + 170

(170,0) => -4 (170) + 170 = -680 + 170 = 510. NO, not equal to 0.

(0,170) => -4 (0) + 170 = 0 + 170 = 170. YES

(12, 126) => -4 (12) + 170= -48 + 170= 122. No, not equal to 126.

(50,30) => -4(50) + 170 = -200 + 170 = 130. No, not equal to 30.

(5, 150) => -4(5) + 170 = -20 + 170 = 150. YES

(60,-70) => -4(60) + 170 = -240 + 170 = -70. YES

4 0

3 years ago

Read 2 more answers

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3 years ago

How is your day? also for the sake of asking, whats 5×1Love yallllll! ​

How is your day? also for the sake of asking, whats 5×1Love yallllll!