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grin007 [14]
3 years ago
6

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi

ned by the equation d = 144 – 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. For which interval of time is Greg’s car moving in the air?
Mathematics
2 answers:
Sophie [7]3 years ago
7 0

Answer:

0<t<3

Step-by-step explanation:

D on Edge

USPshnik [31]3 years ago
6 0
D = 144 - 16t^2
144 - 16t^2 ≥ 0
16(9 - t^2) ≥ 0
9 - t^2 ≥ 0
9 ≥ t^2
t ≥ + or -3

Required interval is 0 ≤ t < 3
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“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor
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Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

C=2\pi r

<u><em>Find the radius of the sphere with a 5.50-meter circumference</em></u>

For C=5.50\ m

assume

\pi =3.14

substitute and solve for r

5.50=2(3.14)r

r=5.50/[2(3.14)]=0.88\ m

<u><em>Find the radius of the sphere with a 7.85-meter circumference</em></u>

For C=7.85\ m

assume

\pi =3.14

substitute and solve for r

7.85=2(3.14)r

r=7.85/[2(3.14)]=1.25\ m

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

SA=4\pi r^{2}

<u><em>Find the surface area of sphere with a 5.50-meter circumference</em></u>

For r=0.88\ m

assume

\pi =3.14

substitute

SA=4(3.14)(0.88)^{2}

SA=9.73\ m^{2}

<u><em>Find the surface area of sphere with a 7.85-meter circumference</em></u>

For r=1.25\ m

assume

\pi =3.14

substitute

SA=4(3.14)(1.25)^{2}

SA=19.63\ m^{2}

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

<u><em>Find the cost of sphere with a 5.50-meter circumference</em></u>

9.73*(92)=\$895.16

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

<u><em>Find the cost of sphere with a 7.85-meter circumference</em></u>

19.63*(92)=\$1,805.96

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

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What is the equation of the line that haas the x-intercept of -5?
Alexus [3.1K]

Answer:

A

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