For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
Yes.
This is the same thing as:
![\frac{36}{6} =6](https://tex.z-dn.net/?f=%20%5Cfrac%7B36%7D%7B6%7D%20%3D6)
How many times can 6 go into 36?
6 times
Try:
6*6=36
So true.
Hope this helps! :D
Answer:
System of equations:
L = 5W + 7
2W + 2L = P
L = 62 cm
W = 11 cm
Step-by-step explanation:
Given the measurements and key words/phrases in the problem, we can set up two different equations that can be used to find both variables, length and width, of the rectangle.
The formula for perimeter of a rectangle is: 2W + 2L = P, where W = width and L = length. We also know that the L is '7 more than five times its width'. This can be written as: L = 5W + 7. Using this expression for the value of 'L', we can use the formula for perimeter and solve for width:
2W + 2(5W + 7) = 146
Distribute: 2W + 10W + 14 = 146
Combine like terms: 12W + 14 = 146
Subtract 14 from both sides: 12W + 14 - 14 = 146 - 14 or 12W = 132
Divide 12 by both sides: 12W/12 = 132/12 or W = 11
Put '11' in for W in the equation for 'L': L = 5(11) + 7 or L = 55 + 7 = 62.
Answer:
ab=de corresponding side of the congruent triangle are equal
5x+6=3x+10
5x-3x=10-6
2x=4
x=4/2=2
2 is your answer