Look carefully at the first pair: (−3, 9), (−3, −5) Note that x does not change, tho' y does. This is how we recognize a vertical line (whose slope is undefined). The equation of this vertical line is x = -3.
Looking at the second pair: from (3,4) to (5,6), x increases by 2 and y by 2; thus, the slope is m = rise/run = 2/2 = 1.
Third pair: as was the case with the first pair, x does not change here, and thus the equation of this (vertical) line is x=0 (which is the y-axis). The slope is undefined.
Answer:
2x^2*(2x^3+8x^2+2x) or
4x^5+16x^4+4x^3
Step-by-step explanation:
A=[2x^3-x+(8x^2+3x)]*4x^2/2=2x^2*(2x^3+8x^2+2x)
or A=4x^5+16x^4+4x^3
Let the smallest integer be x
The second integer = x + 2
The third integer = x + 4
x + x + 2 = x + 4 + 24
2x + 2 = x + 28
2x - x = 28 - 2
x = 26
So the numbers are 26, 28, 30
Answer:
maximum height is 4.058 metres
Time in air = 0.033 second
Step-by-step explanation:
Given that the equation height h
h = -212t^2 + 7t + 4
What is the toy's maximum height?
Let us assume that the equation is a perfect parabola
Time t at Maximum height will be
t = -b/2a
Where b = 7 and a = - 212
t = -7/ - 212 ×2
t = 7/ 424 = 0.0165s
Substitute t in the main equation
h = - 212(7/424)^2 + 7(7/424) + 4
h = - 0.05778 + 0.115567 + 4
h = 4.058 metres
Therefore the maximum height is 4.058 metres
How long is the toy in the air?
The object will go up and return to the ground.
At ground level, h = 0
-212t^2 + 7t + 4 = 0
212t^2 - 7t - 4 = 0
You can factorize the above equation and pick the positive time t since time can't be negative
Or
Since we have assumed that it's a perfect parabola,
Total time in air = (-b/2a) × 2
Time in air = 0.0165 × 2 = 0.033 s
