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3 years ago

Simplify these expressionsa) 7p+6q-2p+q

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

8 0

Answer:

5p + 7q

Step-by-step explanation:

7p + 6q - 2p + q

All you need to do is....

Combine Like Terms

* Put all the ones with "p"'s together, and all the ones with "q"'s together

7p with -2p and 6q with q

7p - 2p + 6q + q

5p + 7q

And you cannot simplify it anymore, because they are not like terms (dont have the same variables)

Hope this helped!

Have a supercalifragilisticexpialidocious day!

marusya05 [52]3 years ago

3 0

Answer:

<h3>Rearrange the terms:</h3>

<u>=</u><u> </u><u> </u><u>5p</u><u>+</u><u>7q</u><u> </u><u> </u> as the answer

Step-by-step explanation:

l hope it helps

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Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

Drupady [299]

Answer:

$V = \frac{\pi^2}{8}$

$V = 1.23245$

Step-by-step explanation:

Given

$y = \cos 2x$

$y = 0; x = 0; x = \frac{\pi}{4}$

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

Apply the following half angle trigonometry identity;

$\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]$

So, we have:

$\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]$

$\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]$

Open bracket

$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)$

So, we have:

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx$

Integrate

$V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0$

Expand

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])$

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])$

$\sin \pi = 0$

So:

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])$

$V = \pi *[{\frac{\pi}{8}]$

$V = \frac{\pi^2}{8}$

$V = \frac{3.14^2}{8}$

$V = 1.23245$

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attashe74 [19]

Answer:

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Step-by-step explanation:

Well the area is Base times Height, so 6 times 4, the squared is just showing that its the area.

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Answer:

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3 years ago

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NISA [10]

Answer:

y=25.5

Step-by-step explanation:

To find y, you need to get it by itself. Ask yourself, what is being done to y, then to undo that, do the opposite. For example, if 3 was added to y, you would need to subtract 3 to cancel out the 3. In this case, what is being done to y? It's being divided by 3. So, you multiply by 3 to cancel that out. Whatever you do to one side of an equation, you also have to do to the other. So,

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Simplify these expressionsa) 7p+6q-2p+q​

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