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Nat2105 [25]
3 years ago
5

The constant of proportionality for this relationship is

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
7 0

Answer: The constant of proportionality is 1.5

Step-by-step explanation:

The equation for constant of proportionality is y=KX. the K stands for the constant of proportionality and since the constant proportionality is 1.5 you will take 1.5 * x and for example x is 2 so you do 1.5 * 2 and you get 3 and 3 is your y. If you plug in the equation you're saying 3 equals 1.5 * 2 which is right.

Charra [1.4K]3 years ago
3 0

Answer:

I'm sorry if this is wrong but:

from 2 and 3 you multiply by 4, from 8 and 12 you divide by 8, 10 and 15 you add 9??

I'm not sure about this at all but I took a guess!

Step-by-step explanation:

Are there any answer choices?

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Find an equation of the line passing through each of the following pairs of points. a (−3, 1), (0, 3)
Svetlanka [38]

\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-3)}}}\implies \cfrac{2}{0+3}\implies \cfrac{2}{3}

\bf \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}{\cfrac{2}{3}}[x-\stackrel{x_1}{(-3)}]\implies y-1=\cfrac{2}{3}(x+3) \\\\\\ y-1=\cfrac{2}{3}x+2\implies y=\cfrac{2}{3}x+3

5 0
3 years ago
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
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leonid [27]

Answer:

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Step-by-step explanation:

4 0
2 years ago
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zheka24 [161]
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5 0
3 years ago
The area of the triangle below is 8.91 square inches. What is the length of the base?<br> 2.2 in
Inessa05 [86]

Answer:

8.1

Step-by-step explanation:

Area of a triangle = ½ × base × height = Area

Rearranged =

area/1/2xheight

The base is the length.

8.91/1/2x2.2

=8.1 inches

4 0
3 years ago
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