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2 years ago

I really need a good explanation on how to solve this, not just the answer please please please..

Mathematics

1 answer:

Alchen [17]2 years ago

4 0

Answer:

Step-by-step explanation:

Mean = add all the numbers up and divide by how many numbers you added

So add (x +120 + 112 + 130 + 128 + 124 ) divided by 6 =118

x is the score you are looking for and 118 is the average he wants

so (x + 614 ) divided by 6 =118

now multiply both sides by 6

x + 614 = 118 times 6

x + 614 = 708

subtract 614 from both sides

x = 94

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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.

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2 years ago

20 POINT QUESTION. EXPLAIN HOW YOU GOT THIS ANSWER. I WILL MARK AS BRAINLIEST THANK YOU :)

Lady_Fox [76]

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3 years ago

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You deposit $7000 into a bank account that has 6% interest compounded SEMIANNUALLY. Determine the amount of money in the bank af

Vlada [557]

Answer:

$11917.03

Step-by-step explanation:

Use Compound Interest Formula:

A = 7,000 (1 + 0.06/2)^2(9)

Simplify:

A = 7,000(1.03)^18

Solve:

A = $11917.03

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2 years ago

a car is traveling at 45 mi/h. write an equation that models the total distance traveled after hours

blsea [12.9K]

45 times number of hours

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3 years ago

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the surface area of a cuboid is 95cm² and its lateral surface area is 63cm². find the area of its base

wel

Answer:

The Area of the base is 16 cm² .

Step-by-step explanation:

Given as :

The surface area of the cuboid = x = 95 cm²

The lateral surface area of the cuboid = y = 63 cm²

Let The Area of the base = z cm²

Now, Let The length of cuboid = l cm

The breadth of cuboid = b cm

The height of cuboid = h cm

According to question

∵ The surface area of the cuboid = 2 ×(length × breadth + breadth × height + height × length)

Or, x = 2 ×(l × b + b × h + h × l)

Or, 95 = 2 ×(l × b + b × h + h × l)

Or, (l × b + b × h + h × l) = $\dfrac{95}{2}$ ....1

Similarly

∵lateral surface area of the cuboid = 2 ×(breadth × height + length × height)

Or, y = 2 ×(b × h + l × h)

Or, 2 ×(b × h + l × h) = 63

Or, (b × h + l × h) = $\dfrac{63}{2}$ ......2

Putting value of eq 2 into eq 1

so, (l × b + $\dfrac{63}{2}$ ) = $\dfrac{95}{2}$

Or, l × b = $\dfrac{95}{2}$ - $\dfrac{63}{2}$

Or, l × b = $\dfrac{95 - 63}{2}$

i.e l × b = $\dfrac{32}{2}$

so, l × b = 16

Now, Again

∵The Area of the base = ( length × breadth ) cm²

So, z = l × b

i.e z = 16 cm²

So, The Area of the base = z = 16 cm²

Hence,The Area of the base is 16 cm² . Answer

5 0

3 years ago