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larisa86 [58]
3 years ago
13

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

ant rate from 20 m/s to a complete stop over a 10 second interval. how does the distance traveled by the truck compare to that of the car?
Mathematics
1 answer:
KIM [24]3 years ago
7 0
To determine the total distance traveled by an object in constant speed, just multiply the speed by the time. So, for the car, the total distance traveled is,
                               distance by car = (20 m/s)(5s) = 100 m
For decelerating object, the distance traveled is,
                          distance = Vt - (at^2)/2
For the truck,
         distance by truck = (20 m/s)(10s) - (2m/s^2)((10s)^2)/2 = 100 m

Relatively, both have traveled the same distance. 
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What are the coordinates of point A?<br> X<br> (8,-6)<br> 0 (-6, 8)<br> (-5, 7)<br> (7-5)
AlexFokin [52]

Answer:

c is the correct one

Step-by-step explanation:

3 0
3 years ago
Solve the linear equation. 14x − 1/2 (4x+6) = 3 (x−4) − 18
LekaFEV [45]

14x - (2x + 3) = 3x - 12 - 18 \\ 14x - 2x - 3 = 3x - 30 \\ 12x - 3 = 3x - 30  \\ 9x = - 27 \\ x =  - 3
7 0
3 years ago
Read 2 more answers
Hi everyone I was wondering if someone could please help me out with this problem and explain it to me
PSYCHO15rus [73]
Join the centre O to the chord (let it be MN) & let OH be the perpendicular to the chord

OH bisects MN into 2 equal parts (each one is x/2)
OMH is a right triangle with one side =8, the second leg =x/2 & the hypotenuse = 12 (Radius)
Apply Pythagoras:

12² = 8² +(x/2)² ==>144=64 + x²/4 ==> x²=4(144-64) =320

x²=320==> x=√320 =17.88 ≈17.9



3 0
3 years ago
(pls help quick and explain how you got the answers for brainliest)
scoray [572]

You only have to apply the theorem of Pythagoras here. Remember the square on the hypotenuse (the longest side) is equal to the sum of the squares on the other two sides :

1. AB is the hypotenuse, so, according to the theorem we can write :

AB² = AC² + CB²

c² = 5² + 4²

c²= 25 + 16

c² = 41

applying the square root of 41 we get :

c ≈ 6.40 rounded to the hundred

The next cases are exactly the same thing so there is no need for explanation :

2.

AB is the hypotenuse here because it is the biggest side clearl :

AB² = AC² + CB²

25² = 15² + b²

Thus

b² = 25² - 15²

we just subtracted 15² on each side of the equation

b² = 625 - 225

b² = 400

applying the square root of 400 we get

b = √400 = 20

So AC = 20

3. The longest side is clearly AB = 60

So

AB² = AC² + CB²

60² = 40² + a²

subtracting 40² on each side of the equation we get :

a² = 60² - 40²

I let you finish this using your calculator and doing exactly like the previous cases

4.

AB is the hypotenuse,

AB² = AC² + CB²

23² = b² + 14²

Subtracting 14² from each side of the equation we get

b² = 23² - 14²

5.

AB is the biggest side :

AB² = AC² + CB²

29² = 23² + a²

We subtract 23² on each sides of the equation :

a² = 29² - 23²

You can finish with your calculator

6.

AB² = AC² + BC²

78² = b² + 30²

subtraction...

b² = 78² - 30²

Good luck :)

7 0
2 years ago
HELP Find the value of x
emmasim [6.3K]
X=100
360-63= 297
297-80=217
217-100=117
117/3=39
4 0
2 years ago
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