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Nataliya [291]
3 years ago
9

During a recent eruption, the volcano spewed out copious amount of ash. One small piece of ash was ejected from the volcano with

initial velocity of 368 ft/sec. The height H, in feet, of our projectile is given by the equation H=-16t^2+368t.
Where t is the time in seconds.
We assume that the volcano has no height
1. When does the ash projectile reach its maxim eight?
2. What is its maxim height?
3. When does the ash projectile return to the ground?
Mathematics
1 answer:
nlexa [21]3 years ago
6 0
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.
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vlada-n [284]

Answer:

18 - (-12)

Step-by-step explanation:

Final temp. - Initial temp. = the change in temperature

By -12, we're indicating that is below zero (which is obvious) meaning that it is not in the "normal" scale nor in the imaginary scale of numbers. So, we can operate the following...

18 - (-12) = 30

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3 years ago
What is the slope-intercept equation of the line shown below? (3,4) (-3,-2)
AysviL [449]
Find the gradient
m = y2-y1 / x2-x1
m = 4-(-2) / 3-(-3)
m = 6 / 6
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4 = 1(3) + c
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Therefore, the equation is:
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4 years ago
Find the area enclosed between f(x)=0.3x2+7 and g(x)=x from x=−4 to x=8.
denis23 [38]
First, we sketch a picture to get a sense of the problem. g(x)=x is a diagonal line through (0,0) with slope = = 1. Since we are interested in the area between x = -4 and x = 8, we find the points on the line at these values. These are (-4, -4) and (8,8).

f(x) is a parabola. It's lowest point occurs when x = 0. It is the point (0,7). At x = -4 and x=8 it has the values 11.8 and 26.2 respectively. That is, it contains the points (-4, 11.8) and (8,26.2).

From these we make a rough sketch (see attachment). This is a sketch and mine is very incorrect when it comes to scale but what matters here is which of the curves is on top, which is below and whether they intersect anywhere in the interval, so my rough sketch is good enough. From the sketch we see that f(x) is always above (greater than) g(x).

To find the area between the curves over the given interval we integrate their difference and since f(x) is strictly greater than g(x) we subtract as follows: f(x) - g(x). The limits of integration are the values -4 and 8 (the x-values between which we are looking for the area.

Now let's integrate:
\int\limits^{8}_ {-4}f(x)-g(x) \, dx = \int\limits^{8}_ {-4}.3 x^{2} +7-x \, dx
The integral yields: [tex](\frac{.3 (8)^{3} }{3} +7(8)- \frac{ (8)^{2} }{2}) -(\frac{.3 (-4)^{3} }{3} +7(-4)- \frac{ (-4)^{2} }{2}) = 117.6 [/tex]
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4 0
3 years ago
The area of a parking lot is 805 square meters. A car requires 5 meters and a bus requires 32 square meters of space. There can
zlopas [31]

Answer:

  80 cars will maximize revenue

Step-by-step explanation:

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Thus the available space should be used to park the maximum number of cars.

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3 years ago
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Reil [10]
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therefore there are left 4/8 pound of trail mix or, what is the same, 1/2 pound of it
4 0
4 years ago
Read 2 more answers
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