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bija089 [108]
3 years ago
10

5. Which third-party wants little to no government interference?

Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

5. Libertarians.

6. Liberals.

7. Republicans.

8. Democrats.

Step-by-step explanation:

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Is the following shape a rectangle
Gelneren [198K]

Answer:

c

Step-by-step explanation:

no because the adjacent sides are not perpendicular.

4 0
3 years ago
Please help!!!!!!!!!!!!!!!
babunello [35]

Answer:

Answer A

Step-by-step explanation:

Answer A indicates some minimal value for the set of points to the left (negative side of the distribution) which are associated with negative skewness since the mean is at 22 also with larger deviation to the left shown in the median of the points to the left further to the negative side of the plot.

Answers B and C show symmetric distributions (no skewness at all).

Answer D would represent skewness to the right (positively skewed)

So the correct answer is answer A.

8 0
3 years ago
If a builder needs p pieces of wood for every 5s square feet of the building, what is the total amount of lumber of he needs to
Black_prince [1.1K]

Answer: He will need 700 pieces of wood.

Step-by-step explanation: Divide the total by how many square feet one piece of wood will cover, so 3,500/5, and you'll get 700.

Hope this helps.

5 0
2 years ago
Which system below has no solution?
Alex787 [66]

Answer:

Step-by-step explanation:

A system has no solution if the 2 (or more) functions that make up the system do not intersect.  The only pair of functions here that do not have an intersection is found in D. You could graph these 2 on a graphing calculator to see that this is true.

4 0
3 years ago
Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
3 0
3 years ago
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