Does anyone know how to solve this?

First, find the x-intercept of g(x) = x³ - 1. x³ = 1, so x = 1. The x-intercept is (1,0).

We must now find the equation of the tangent line to the graph at (1,0):

slope of tangent line = 3x^2 with x = 1, or m = 3(1)^2 = 3.

Last, find the equation of the line with slope 3 that passes through (1,0):

0 = 3(1) + b. Then b = -3, and the equation of the tangent line at (1,0) is

y = 3x - 3.

You can find the equation of a line if you know the slope of the line and one point on the line using the point-slope form of the equation of a line.

First, let's find a point on the line. We will find the x-intercept. At the x-intercept, y = 0, so let y = 0 and solve for x.

g(x) = x^3 - 1

x^3 - 1 = 0

x^3 = 1

x = 1

The value of x at the x-intercept is 1, so the x-intercept is point (1, 0). We have the point we need.

Now we need the slope. The slope of this function at any point x is given by 3x^2. Let's find the slope at the x-intercept. At the x-intercept, x = 1, so the slope is

m = 3x^2 = 3(1)^2 = 3

We know the tangent line at the x-intercept has slope 3 and passes through the point (1, 0). Now we use the point-slope form of the equation of a line to find the equation of the tangent.

y = 3x - 3

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl