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aivan3 [116]
2 years ago
7

Does this scatter plot show a positive trend a negative trend or no trend​

Mathematics
1 answer:
maw [93]2 years ago
6 0

The dots are moving in a downward direction from left to right, so it is a negative trend.

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An organic farmer wanted to find out how many eggs a typical hen lays during 1 month. She chose 8 hens at random from the 60 hen
Marat540 [252]
First, you add them all together, and divide them by the number of quantities there are, so:
9+9+8+11+11+12+12+8= <span>80. Then you divide it by 8 (how many numbers there are) to get 10!</span>
8 0
3 years ago
Read 2 more answers
Please help me out here i need a real answer pleasseeee
insens350 [35]

Answer:

33

Step-by-step explanation:

35*3-8+33=33

8 0
3 years ago
Hello pls help me guy but pls no random links
lbvjy [14]

Answer:  Choice A

Work Shown:

5% of $25 = 0.05*25 = 1.25

4 0
2 years ago
What is the area of a square whose<br> diagonal is 14cm
xenn [34]

Answer:

98 cm^{2}

Step-by-step explanation:

The formula for the area of a square when given the diagonal is:

A = \frac{1}{2} d^{2} Where d is the value of the diagonal so it becomes:

0.5 × 14 × 14

This is equal to 98 cm^{2}.

HOPE THIS HELPED

7 0
2 years ago
Calculus 2 master needed; evaluate the integral PLEASE SHOW STEPS IF IM WRONG <img src="https://tex.z-dn.net/?f=%5Cint%7Bsin%5E3
sweet [91]

Answer:

Yes, you answer is correct! It just needs to be simplified :)

Step-by-step explanation:

So we have the integral:

\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx

As you had done, we can split off the numerator:

=\int \frac{\sin(x)(\sin^2(x))}{\sqrt{\cos(x)}}dx

Using the Pythagorean Identity, this is:

=\int \frac{\sin(x)(1-\cos^2(x))}{\sqrt{\cos(x)}}dx

Now, we can do u-substitution. Let u equal cos(x). Thus:

u=\cos(x)\\du=-\sin(x)dx\\-du=\sin(x)dx

So:

=\int \frac{1-u^2}{\sqrt{u}}(-du)

Simplify:

=-\int\frac{1-u^2}{\sqrt u}du

We can then split the terms:

=-\int \frac{1}{\sqrt u}-\frac{u^2}{\sqrt u}du

Expand the integral:

=-(\int \frac{1}{\sqrt u}du-\int\frac{u^2}{\sqrt u}du)

Simplify each of the u.

For the left, that is simply u^-1/2.

For the right, it is u^(2-1/2) or u^3/2. Thus:

=-(\int u^{-\frac{1}{2}}du-\int u^{\frac{3}{2}}du)

Reverse Power Rule:

=-(\frac{u^{1+-\frac{1}{2}}}{1+-\frac{1}{2}}-\frac{u^{1+\frac{3}{2}}}{1+\frac{3}{2}})

Simplify:

=-(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{5}{2}}}{\frac{5}{2}})

Simplify further:

=-(2u^{\frac{1}{2}}-\frac{2u^{\frac{5}{2}}}{5})

Distribute the negative:

=-2u^{\frac{1}{2}}+\frac{2u^{\frac{5}{2}}}{5}

And substitute back cos(x) for u:

=-2\cos^{\frac{1}{2}}(x)+\frac{2\cos^{\frac{5}{2}}(x)}{5}

And this is precisely what you got, so well done!

We can simplify this by first multiplying the first term by 5 to get a common denominator. So:

=-\frac{10\cos^{\frac{1}{2}}(x)}{5}+\frac{2\cos^{\frac{5}{2}}(x)}{5}

Combine:

=\frac{-10\cos^{\frac{1}{2}}(x)+2\cos^{\frac{5}{2}}(x)}{5}

Factor out a cos^(1/2)(x) and a 2. Since we factored out a cos^(1/2)(x), we need to subtract their exponents inside. Thus:

=\frac{2\cos^{\frac{1}{2}}(x)(-5\cos^{\frac{1}{2}-\frac{1}{2}}(x)+\cos^{\frac{5}{2}-\frac{1}{2}}(x))}{5}

Simplify:

=\frac{2\cos^{\frac{1}{2}}(x)(-5+\cos^2(x))}{5}

Simplify:

=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}

And, of course, C:

=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C

So:

\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C

And we're done :)

3 0
3 years ago
Read 2 more answers
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