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Lesechka [4]
2 years ago
7

Need help on this one please

Mathematics
1 answer:
lesya [120]2 years ago
3 0

Answer: y = -2x - 4

Step-by-step explanation:

This question asks that you give your answer in slope-intercept form, which is written as y = mx + b, where m is the slope, and b is the y-intercept.


In order to write an equation for this line in slope-intercept form, you need to find the slope and the y-intercept. You can find the slope by selecting any two clear points on the graph and finding the [change in y] ÷ [change in x].

For example, I’ll use (-3, 2) and (-2, 0):

Change in y: 2 - 0 = 2

Change in x: -3 - (-2) = -1

Change in y ÷ Change in x = 2 ÷ (-1) = -2


So, the slope, m, is -2. Finding b, or the y-intercept is a lot more straightforward; you just find where the line intercepts the y-axis, or when x = 0. Looking at the graph, we can see that the y-intercept is -4, so b = -4.


Now, we can put these together in an equation.

y = mx + b, where m = -2 and b = -4


y = -2x -4

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Step-by-step explanation:

1a) use letter "P" for perimeter

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What tests are used to determine the radius of convergence of a power series? A. Divergence TestB. Root Test C. Comparison Test
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Help me with these 3 questions, please
Zina [86]

Answer:

1. x = -4y ---> y = (-1/4)x

slope = -1/4. y-intercept = (0,0)

2. y = -2x + 4

3. y = (1/3)x - 1

Step-by-step explanation:

1. Re-write your equation so that x is on the right and y is on the left:

x = -4y ---> y = (-1/4)x

slope = -1/4. y-intercept = (0,0)

2. y-intercept = (0,4) ----> P1

x-intercrpt = (2,0) ----> P2

slope m = (y2 - y1) / (x2 - x1)

= (0 - 4)/(2 - 0)

= -2

therefore, y - y1 = mx - x1 ---> y - 4 = -2x

or y = -2x + 4

3. y-intercept = (0,-1)

x-intercept = (3,0)

m = (0 - (-1)) / (3 -0) = 1/3

y - (-1) = (1/3)x - 0 ---> y = (1/3)x - 1

5 0
3 years ago
(b) dy/dx = (x - y+ 1)^2
Elanso [62]

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

Then the resulting ODE in v(x) is separable, with

1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx

On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

3 0
3 years ago
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