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blagie [28]
3 years ago
11

Y>-3x+1 How do I graph this

Mathematics
1 answer:
Brut [27]3 years ago
4 0

Answer:

<u>The graph of y>-3x+1 through the points </u><u>(0,1)</u><u> and </u><u>(1,-2)</u>

Step-by-step explanation:

y=3x+1 is in the slope intercept form

y=mx+b

m = the slope in this case 3

b = the y intercept in this case 1.

Think of b the y intercept as the beginning this is where to start the graph.

So start with (0,1) the y intercept and make the first point at (0,1)

Think of m ( the slope) as the mountain slope, This is the angle of the slope of the line

So in this case go up three on the y axis and over 1 on the x axis.

make the next point at 1 + 3 = 4 for the y value and 0 + 1 = 1 for the x value. Make the next point at ( 1, 4)

Now connect the points. creating the line

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Which equation gives the lines show on Graph y= 2x+4 <br> y= -1/2x-4<br> y=1/2x+4 <br> y=2x-4
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The answer is the second one y=1/2x+4
5 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=
BARSIC [14]
<h3>Answer:  -2</h3>

======================================================

Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0
3 years ago
A summarine is only allowed to change his depth by rising Toward the surface in 50 m stages it starts off at negative 235 m how
Vladimir [108]

Answer:

It will take 5 stages to reach the surface.

Step-by-step explanation:

-225/50=4.5 or what times 4.5 equals -225

Hope this helps.

3 0
2 years ago
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