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Ratling [72]
3 years ago
9

Please help, WILL MARK BRILLIANT! Is the relationship shown by the data linear? If so, model the data with an equation

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0
Hi...the answer is b.. I would recommend plot graph and find the gradient.... then use the point formula
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The equation of a circle is ​ (x−11)2+(y+15)2=144 ​ .<br><br> What is the circle's radius?
yuradex [85]
The equation of a circle is (x-h)^2+(y-k)^2=r^2, where (h,k) is the center and r is the radius.
To find the radius, we would find the square root of 144, which is 12.
r=12

:)
3 0
3 years ago
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Factor completely 3x − 12.<br><br> 3(x − 4)<br><br> 3(x + 4)<br><br> 3x(−12)<br><br> Prime
garri49 [273]
3x - 12
3(3x/3 - 12/3)
3(x - 4)

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2 years ago
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What is the range of the following data set?​
sashaice [31]

Answer:  The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

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2 years ago
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
2 years ago
Jordan is another employee at the same coffee shop. He has worked there longer than Gabe and earns $3 more per hour than Gabe. C
ANEK [815]

Answer:

I think we need to see something

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2 years ago
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