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stiks02 [169]
3 years ago
11

4(28) = 4(20) + 4(?)

Mathematics
1 answer:
brilliants [131]3 years ago
4 0
4(28) = 4(20) + 4(8)
The answer is 8
Because 28 is 20 and 8
You might be interested in
Can two positive integers have their H.C.F and L.C.M as 12 and 512 respectively? justify.
DaniilM [7]

Here is your answer

NO

Reason:

H.C.F. divides the L.C.M. completely.

Here 512 is not completely divided by 12

So, two numbers can't have their H.C.F and L.C.M as 12 and 512 respectively.

HOPE IT IS USEFUL

7 0
3 years ago
Read 2 more answers
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)

5 0
3 years ago
IF:
castortr0y [4]
So they go up and first they double then triple etc do the answer would be 3050
4 0
4 years ago
Sorry for the quality, if anyone could help. It’s geometry.
user100 [1]

Answer:

5

Step-by-step explanation:

8 0
3 years ago
Please help I will mark Brainlest
alexandr1967 [171]

Answer:

Step-by-step explanation:

ES 100 CHAOOOOOOOOOOOOOOOOOOOOOOOO

3 0
3 years ago
Read 2 more answers
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