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3 years ago

Which of the following could be the graph of the line y = 3x - 2?

Mathematics

2 answers:

Travka [436]3 years ago

8 0

Since there is no picture provided, I can't give you the answer, but I can help you solve it.

To solve this you have to first look at the b part of the formula y=mx+b, so -2.

Then, you can find -2 on the y-axis (the vertical axis)

and then you could move upwards 3 spaces, then right 1 space.

You would do this as the slope is 3/1 (the mx part of the equation).

Finally, you continue to move up 3, and right 1 until there are enough points on the graph to make a line, and that will help you see what graph is correct.

It should look something like this:

morpeh [17]3 years ago

5 0

You didn't give a graph but I will visualize what it looks like for you.

The line is going through (0,-2) because that is the y-intercept.

The line has a slope of 3.

So, the line goes through:

(1,1)

(2,4)

(3,7)

(4,10)

(5,13)

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3 years ago

Which expression is equivalent to > 7^8/7 7^9 7^8 7^7 1/7^8

tankabanditka [31]

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2 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

State the conditions that the following graph meets. (Enter the equation or inequality that is shown by the graph.)

shutvik [7]

The graph shows a horizontal line intersecting (0, 2).

Your answer is: y = 2.

7 0

3 years ago

Read 2 more answers

Which of the following functions is graphed below.

Valentin [98]

Answer: Option A

$y=\left \{ {{x^2 +2;\ \ x$

Step-by-step explanation:

In the graph we have a piecewise function composed of a parabola and a line.

The parabola has the vertex in the point (0, 2) and cuts the y-axis in y = 2.

The equation of this parabola is $y = x ^ 2 +2$

Then we have an equation line $y = -x + 2
$

Note that the interval in which the parabola is defined is from -∞ to x = 1. Note that the parabola does not include the point x = 1 because it is marked with an empty circle " о ."

(this is $x< 1$ )

Then the equation of the line goes from x = 1 to ∞ . In this case, the line includes x = 1 because the point at the end of the line is represented by a full circle .

(this is $x\geq 1$ )

Then the function is:

$y=\left \{ {{x^2 +2;\ \ x$

7 0

3 years ago