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2 years ago

April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the

Mathematics

1 answer:

zaharov [31]2 years ago

6 0

Answer:

The maximum height of the arrow is 125 feet.

Step-by-step explanation:

The pathway of the arrow can be represented by the equation,

$h = -16t^2 +80t + 25$ .....(1)

Where h is height in in feet and t is time in seconds.

It is required to find the maximum height of the arrow. For maximum height, $\dfrac{dh}{dt}=0$ .

So,

$\dfrac{d(-16t^2 +80t + 25)}{dt}=0\\\\-32t+80=0\\\\t=\dfrac{80}{32}\\\\t=2.5\ s$

Put t = 2.5 s in equation (1). So,

$h = -16(2.5)^2 +80(2.5) + 25\\\\h=125\ \text{feet}$

So, the maximum height of the arrow is 125 feet.

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WILL give BRAINLIEST

Nostrana [21]

Answer:

$560.00

Step-by-step explanation:

Let b represent Tony's average daily balance. The finance charge is computed from ...

... finance charge = 2.25% × b

We can fill in the given value and divide by the coefficient of b.

... $12.60 = 0.0225b

... $12.60/0.0225 = b = $560.00

Tony's average daily balance was $560.00.

3 0

3 years ago

Molly buys 12 rolls of wallpaper for £111.00.<br> How much is 1 roll of wallpaper?

alex41 [277]

Answer: £9.25

111.00/12=9.25

3 0

3 years ago

Read 2 more answers

Use the diagram to determine which statement is true

hichkok12 [17]

A. Area of ABCD - Area of DGA = Area of DEFG
s^2 - 1/2bh = s^2
(5)^2 - 1/2(4)(3) = (3)^2
25 - 1/2(12) = 9
25 - 24 = 9
1 not equal to 9

B. Area of ABCD - Area of GHIA = Area of DGA
s^2 - s^2 = 1/2bh
(5)^2 - (4)^2 = 1/2(4)(3)
25 - 16 = 1/2(12)
9 not equal to 6

C. Area of ABCD + Area of DGA = Area of GHIA
s^2 + 1/2bh = s^2
(5)^2 + 1/2(4)(3) = (4)^2
25 + 1/2(12) = 16
25 + 6 = 16
31 not equal to 16

D. Area of DEFG + Area of GHIA = Area of ABCD
s^2 + s^2 = s^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25

The answer is D.

5 0

2 years ago

Read 2 more answers

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max

siniylev [52]

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$ ( $y=f(x)$ ).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$