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Ivanshal [37]
3 years ago
11

100 points and brainliest to whoever gives a good explanation and answer to the question. The graph shows a proportional relatio

nship which equation matches the graph?​

Mathematics
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

B

Step-by-step explanation:

We are given a proportional relationship and we want to find the equation that models it.

Since the relationship is proportional, we know that it must go through the origin point (0, 0).

Also, we can see from the graph that it passes through the point (1, 3).

We can now find the constant of proportionality or the slope using the slope formula:

\displaystyle m=\frac{y_2-y_1}{x_2-x_1}

Let (0, 0) be (x₁, y₁) and let (1, 3) be (x₂, y₂). Hence:

\displaystyle m=\frac{3-0}{1-0}=\frac{3}{1} = 3

Our constant of proportionality is 3.

A proportional relationship is given by:

y=kx

Where k is the constant of proportionality.

So:

y=3x

Our answer is B.

san4es73 [151]3 years ago
5 0

Answer:

It would be B

Step-by-step explanation:

You can check each question by subsituting the x and y with points on the graph.

Example: (1,3)

y = 3x

3 = 1 x 3

3 = 3

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Pleaseeee helpppppppppp
o-na [289]

Answer:

25 ft^2

Step-by-step explanation:

In direct variation, if y varies directly with x, then the equation has the form

y = kx,

where k is the constant of proportionality. y is proportional to x.

Let's call the area y and the distance x.

Here, the area varies with the square of the distance, so the equation has the form

y = kx^2

Here, y is proportional to the square of x.

We can find the value of k by using the given information.

y = kx^2

When x = 20 ft, y = 16 ft^2.

16 = k(20^2)

k = 16/400

k = 1/25

The equation of the relation is:

y = (1/25)x^2

Now we use the equation we found to answer the question.

What is y (the area) when x (the distance) is 25 ft?

y = (1/25)x^2

y = (1/25)(25^2)

y = 25

Answer: 25 ft^2

5 0
2 years ago
At the supermarket , 6 mangoes cost 6.12
Grace [21]

Answer:

good to know. but what's the question?

4 0
3 years ago
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Lynna [10]
Im not sure what this is, or how to answer it
4 0
3 years ago
The points A(12,7), B(12,−3), C(−2,−3), and D(−2,7) form rectangle ABCD. Which point is halfway between A and B? CLEAR CHECK (12
Flura [38]
We have that
coordinates  are
<span>A(12,7), B(12,−3), C(−2,−3), and D(−2,7) 
</span>
i use the formula of midpoint <span>between A and B
</span>Xm=(x1+x2)/2    Ym=(y1+y2)/2
A(12,7) B(12,−3)
Xm=(12+12)/2=12
Ym=(7+(-3))/2=2

the point  (12,2)   <span>is halfway between A and B</span>
4 0
3 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
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