The size of the second application given the size of the first application and the expression ( x - 3.45 mb) for the size of the second application is 293.55 MB.
<h3>Equation</h3>
Let
- Size of the first application = x
- Size of the second application= x - 3.45 mb
For instance,
if the size of the first application is 297 MB
Size of the second application= x - 3.45 mb
= 297 MB - 3.45 MB
= 293.55 MB
Therefore, the size of the second application given the size of the first application and the expression for the size of the second application is 293.55 MB
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Answer:
/2 and /4 because angle 1 is 90
Sure! quick question. What type of videos do you make? Like the genre.
<span>So, L*W=A Because it is 4 cm longer, L=W+4 Because the area is 96, LW=96 Substitute to get W(W+4)=96 Multiply it out. W^2+4W-96=0 By solving the quadratic, W+12(W-8)=0 so either W+12 or W-8 is zero. The width must be positive, so the width is 8. Therefore the length is 12.
Hope this helps.</span>
Answer:
a) No. t < 0 is not part of the useful domain of the function
b) 2.0 seconds
Step-by-step explanation:
a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.
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b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.
h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0
0 = -4(4t² -7t -2) = -4(4t +1)(t -2)
This has t-intercepts where the factors are zero, at t=-1/4 and t=2.
The ball will hit the ground after 2 seconds.
