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Alina [70]
3 years ago
8

Using 8 coins, how can you make change for 65 cents that will not make change for a quarter?

Mathematics
2 answers:
Ksivusya [100]3 years ago
6 0
<span>2 quarters, 1 dime and 5 pennies

</span>
NemiM [27]3 years ago
5 0
2 dimes, five nickels, 
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The function g(x)=f(x+3)+2. Which of the following is true about the graph of g(x)?
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Answer:

Your Answer is B.

Step-by-step explanation:

If the number inside the () is positive, it shifts left. If it is negative, It shifts right. The other number is up if positive, down if negative.

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Which of the following equations is quadratic?
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Find the Net Pay: 40 hours worked, $7.85 an hour. Deductions: Federal tax $32.24, Social Security $24.02, Other $24.53 Answer $3
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A hungry college student just finished eating a large cheese and pepperoni pizza from​ Doug's Pizza Palace and felt there​ wasn'
maw [93]

Answer:  14

Step-by-step explanation:

  • Degrees of freedom is the number of values in the evaluation of a test statistic that are free to vary.

For t-statistic , the degree of freedom for a sample with sample size 'n' is given by :-

df=n-1

Given : Sample size : n-1

Then, the number of degrees of freedom the t-value will have :-

df=15-1=14

Hence, the t-value will have 14 degrees of freedom.

7 0
3 years ago
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
3 years ago
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