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Nookie1986 [14]
3 years ago
14

A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The h

eight of the airplane above ground after t seconds is given by the equation h=-16t^2+56t+1, where h is the height of the airplane in feet and t is the time in seconds after it is launched. Approximately how long does it take the airplane to reach its maximum height?
A.

0.3 seconds
B.

1.8 seconds
C.

3.5 seconds
D.

6.9 seconds
Mathematics
2 answers:
morpeh [17]3 years ago
7 0

Answer:

The correct answer is option B. 1.8 seconds. I just took the test.


nata0808 [166]3 years ago
3 0
For a vertical projectile, the equation of motion is expressed h(t)=g+bt+c, where g is gravity, b is the initial upward velocity, and c is the starting height. The maxima of this equation is given by -b/2g, which in this case would be -56/-32, or 1.75 seconds after launch
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For f(x) = 0.01(2)^x, find the average rate of change from x = 3 to x = 8.
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8 - 3 = 5
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3 years ago
Please walk me through this​
FinnZ [79.3K]

Answer:

X is 160. because you can add all the value of x. But first substrate 540 with 100 which is 440. then 1/2x + 1/2x+ (x-15) + (x-25) which is further simplified as x + 2x - 40. then it's further simplified as 3x - 40 = 440. then add 40 with 440 which is 480 then divide it by 3 which is 160. therefore X= 160.

3 0
3 years ago
The slope intercept form of the equation of a line that passes through (-2,13) is y =5x-3what is the point slope form of the equ
Zigmanuir [339]

Answer:

y - 13 = 5(x + 2)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Given y = 5x - 3 in  slope- intercept form

with slope m = 5, then

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y - 13 = 5(x - (- 2)), that is

y - 13 = 5(x + 2) ← in point- slope form

6 0
3 years ago
​Joe's annual income has been increasing each year by the same dollar amount. The first year his income was ​$17 comma 90017,900
Vedmedyk [2.9K]

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points (x_1,y_1) and (x_2,y_2), then the equation of line is

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of line is

y-17900=\frac{20300-17900}{4-1}(x-1)

y-17900=\frac{2400}{3}(x-1)

y-17900=800(x-1)

y-17900=800x-800

Add 17900 on both sides.

y=800x-800+17900

y=800x+17100

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

30700=800x+17100

Subtract 17100 from both sides.

30700-17100=800x

13600=800x

Divide both sides by 800.

\frac{13600}{800}=x

17=x

Therefore, in 17th year his income was $30,700.

5 0
3 years ago
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ivolga24 [154]
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