Answer:
Bro just tell them the answer Not lecuter them The answer is C
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
In the composition of <em>f</em><em> </em>and <em>g</em>, the equation is telling us to place the function <em>f</em> inside <em>g</em>. Observing in B that <em>f</em> is (x + 4), then this can be put in place of the x in 3x^2, making it 3(x + 4)^2 and finally the minus 5.
An x-intercept is the point where the function passes the x axis at y=0.
The y-intercept is the point where the function crosses the y axis at x=0
1. It is an x-intercept, so y = 0. The ordered pair would be (-6, 0)
2. It is a y-intercept, so x = 0. The ordered pair would be (0, -2.3)
3. It is a y-intercept, so x=0. The ordered pair would be (0, 3/4)
To find the x-intercept, set y = 0. To find the y-intercept, set x=0
4. y-intercept: y = 3(0) -9. y = -9 The y-intercept is at (0, -9)
x-intercept: 0 = 3x -9. 9 = 3x. x = 3. The x-intercept is at (3, 0)
5. x intercept: 0 = 5x +10. -10 = 5x. x = -2. The x-intercept is at (-2, 0)
y-intercept: y = 5(0) + 10. y = 10. The y-intercept is at (0, 10)
If you look at finding the y-intercepts in the two problems above, you may see there is a pattern forming. The y-intercept is the number that your adding or subtracting that is located after the x (ex. y = 4x - 2 - the y-intercept would be -2)
9. First, find the x and y-intercepts. The y intercept is -3 You’d graph that at (0, -3). 0 = -1/2x - 3. -1/2x = 3. x = -6 so the x-intercept is at (-6, 0). If you only need to graph 2 points, then you can graph just those two points and draw a line between them.
To graph y=-1/2x + 3, start at the y-intercept and use the slope (-1/2) to find other points. Because your slope is -1/2, you’d go down 1 unit and then to the right 2 units. That would be your next point. If you wanted your line to go further up, go up one unit and then to the left 2 units. That would be your next point.
I am not sure what you need to do on 11 and 12
I think you should try 7, 8 and 10 on your own and let me know if you have any questions on them or if you are stuck on anything.
