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Shtirlitz [24]
1 year ago
13

The volume of a prism is the product of its height and area of its base, V = Bh. A rectangular prism has a volume of 16y4 + 16y3

+ 48y2 cubic units. Which could be the base area and height of the prism? a base area of 4y square units and height of 4y2 + 4y + 12 units a base area of 8y2 square units and height of y2 + 2y + 4 units a base area of 12y square units and height of 4y2 + 4y + 36 units a base area of 16y2 square units and height of y2 + y + 3 units
Mathematics
2 answers:
Komok [63]1 year ago
6 0
base 16y^2
height y^2 + y + 3
V = b*h
V = 16y^2(y^2 + y + 3)
V = 16y^4 + 16y^3 + 48y^2
Last option
Sliva [168]1 year ago
3 0

Answer:

<u>D! </u>

<u>a base area of 16y2 square units and height of y2 + y + 3 units</u>

<u></u>

<u />

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Answer:

Initial population of Rabbit = 5 rabbit

After 2 months

Population of Rabbit = 10

After 4 months

population of rabbit = 20

Formula for growth is :

G = G_{0}[1 + R]^n, where G is final population and G_{0} is initial population, and R is growth rate.

1. 10 = 5 [1 +R]²

Dividing both sides by 5 , we get

2 = (1 + R)²

→ R + 1 = √2  ⇒ taking positive root of 2

→R = √2 -1

Amount of rabbit after 1 year = 20(1 + \sqrt2 -1)^8= 20 \times (\sqrt2)^8= 20 \times 2^4= 20 \times 16= 320




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10 x 62 is the equation for this problem because every row has 62 and there are 10 rows.
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Jenny is taking a vacation to Florida. She travels 70 kilometers per hour for 2 hours, and 63 kilometers per hour for 5 hours. O
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Jenny traveled 70 km/h over 2 hours and 63 km/h over 5 hours. Her travel time was a total of 7 hours. 

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6 0
2 years ago
You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in
emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

j(t)=95t-9t^2

j'(t)=95-18t

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

t=\frac{95}{18}

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

j(5.28)=95(5.28)-9(5.28)^{2}

j(5.28)=250.69ft

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

j(t)=95t-9t^{2}

j'(t)=95-18t

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3)  yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

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