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krok68 [10]
3 years ago
7

Is Y equals 2.5 X proportional

Mathematics
1 answer:
Ivahew [28]3 years ago
6 0

Answer:

no but 1+1=2 hehe

Step-by-step explanation:

follow me on tt: dxddy.drip0

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15X^2+20x factor greatest common monomial
velikii [3]
Both 15 and 20 can be divided evenly by 5, so 5 is the greatest common monomial:  5(3x^2 + 4x)  or  5x(3x + 4) 
4 0
3 years ago
The sum of the components of anything equals the whole thing. Which property/postulate does this statement represent?
IgorC [24]
The property that is being described in the statement "The sum of the components of anything equals the whole thing" would be the Partition Postulate. It is simply the whole is equal to the sum of its parts. For instance we have a line where it contains points W, X, Y and Z, then WX + XY + YZ = WZ.
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3 years ago
Please help me stats
Korolek [52]
The answer is A.....
5 0
3 years ago
What is the equation in point slope form of the line that is parallel to the given line and passes through the point (4,1)?
vovikov84 [41]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

y-1=\stackrel{\stackrel{m}{\downarrow }}{2}(x-4)\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}

so we're really looking for the equation of a line with a slope of 2 and that it passes through (4 , 1)

(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{2}(x-\stackrel{x_1}{4})

kinda twins.

8 0
1 year ago
Evaluate the line integral, where C is the given curve. C sin(x) dx + cos(y) dy, where C consists of the top half of the circle
kenny6666 [7]

Parameterize the circular part of C (call it C_1) by

x=4\cos t

y=4\sin t

wih 0\le t\le\pi, and the linear part (call it C_2) by

x=-4-t

y=4t

with 0\le t\le1.

Then

\displaystyle\int_C\sin x\,\mathrm dx+\cos y\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}\sin x\,\mathrm dx+\cos y\,\mathrm dy

=\displaystyle\int_0^\pi(-4\sin t\sin(4\cos t)+4\cos t\cos(4\sin t))\,\mathrm dt+\int_0^1(-\sin(-4-t)+\cos4t)\,\mathrm dt

=0+\displaystyle\int_0^1(\sin(t+4)+\cos4t)\,\mathrm dt

=\cos4-\cos5+\dfrac{\sin4}4

7 0
3 years ago
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