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2 years ago

8

Graph the lines in standard form 2x+y=9

2 answers:

Nezavi [6.7K]2 years ago

8 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

A graphing calculator produces a graph very nicely.

__

Setting x=0, you have the equation y=9, so you know the y-intercept.

Setting y=0, you have the equation 2x=9, so you know the x-intercept is 9/2 = 4.5. It can be easier to set y=1 to find the corresponding point x=(9-1)/2 = 4.

Through the points (0, 9) and (4, 1), you can draw the line representing the relation.

julsineya [31]2 years ago

5 0

Answer:

to graph this all you need to do is plot the y-intercerpt which is 9 in this case. then use the slop (-2) which means go two down from the y intercept and one to the right. all you need to do then is connect the points with a line.

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Could somebody help me understand this?

pantera1 [17]

Answer:

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Step-by-step explanation:

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1 year ago

45 POINTS PLEASE HELP!

andrew-mc [135]

Hello,

Question:

Jack popped some popcorn, but 3 of the 150 kernels did not pop.

Determine which popcorn snacks below has the same ratio of unpopped to total kernels as Jack's popcorn.

We Know:

3 out of 150 did not pop

Solution:

3/150 = 3 to 150

Find ratios similar to this one,

4/200

2/100

1/50

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Answers:

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8 0

3 years ago

Read 2 more answers

Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx

shutvik [7]

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$ , so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$ :

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

$\csc v-\cot v=x+C$

$\csc(x+y)-\cot(x+y)=x+C$

Given that $y(0)=\dfrac\pi2$ , we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

$\csc(x+y)-\cot(x+y)=x+1$

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