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3 years ago

Help me with any of this

Mathematics

2 answers:

Inessa [10]3 years ago

7 0

Number 4 is 31. number 7 is 1. number 8 is 21. number 9 is 64. number 10 is 29. number 12 is 38. vote me brainliest please

JulsSmile [24]3 years ago

5 0

1. (-50) - 46 = -96 [add, then get the bigger number that is positive or negative.]
2. 28 - (-3) = 31 [ KCC (Keep, Change, Flip), then add.]
3. 25 - 24 = 1 [just subtract.]

You should know how to the rest.

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What is the y-intercept of y = 5? Choose 1 answer: 5,5 5,0 0,5 0,0

evablogger [386]

Answer:

y-intercept is at (0, 5) for y = 5

Step-by-step explanation:

for y = 5 is a horizontal line

7 0

3 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:

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