How do I solve this ?

Can you please use my for gauth code GHD989

viva [34]

Answer:

what........???????............

4 0

2 years ago

An exam consists of two parts, Section X and Section Y. There can be a maximum of 95 questions. There must be at least 20 more q

GuDViN [60]

Answer:

X + Y <= 95.

Y >= 20+ X.

Step-by-step explanation:

It is given that section X and section Y can have hardly 95 questions.

Hence either both have them has total 95 questions or less than 95 questions.

If we take X as the number of questions section X has and Y as the number of questions section Y has then X + Y <= 95.

Also it is given that Y has at least more 20 more questions than Y has. Hence Y can have 20 more question than X has or more than 20 more question than X.

That is Y >= 20+ X.

As shown in the graph the line passing through (95, 0) and (0, 95) the arrows are towards the origin because the origin satisfies the equation.

In the other line the origin that is (0, 0) does not satisfy the equation. Hence the arrows will not be towards the origin.

4 0

3 years ago

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:

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

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

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

or

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

$V = 1.23245$

4 0

3 years ago

Find the area of the shaded region 6cm 6cm

Mashutka [201]

Hope this helps! the answer is option (A). if you still don’t understand after looking at my solution feel free to clarify ya

4 0

3 years ago

Roni and Allie are mowing the grass at the soccer field. Roni has a riding lawn mower and can mow the field in 30 minutes. Allie

KIM [24]

<span>Q Roni and Allie are mowing the grass at the soccer field. Roni has a riding lawn mower and can mow the field in 30 minutes. Allie is pushing a lawn mower and can mow the field in 75 minutes.
If Roni and Allie work together to mow the field, what part of the field would Roni mow?

A He would mow the bigger part of the field
</span>

7 0

3 years ago

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