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icang [17]
3 years ago
11

If f(x) = x2 - 3x + 1 and f(2 a) = 2f(a) then a is equal to​

Physics
1 answer:
loris [4]3 years ago
5 0

For these question, it has two separate equations: 2f(a) and f(2a) .

For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation

For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.

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A man pulls on his dogs leash to keep him from running after the bicycle. Which term best describes this example?
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What term do you mean? like what he did to the dog is he stopped the dog
5 0
3 years ago
Read 2 more answers
A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)
STALIN [3.7K]

The car traverses a distance x after time t according to

x=\dfrac12at^2

where a is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity v after 2 s of travel is given by

v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}

which makes C the most likely answer.

3 0
4 years ago
A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe
Allisa [31]

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

8 0
3 years ago
What types of collisions can result from making unsafe passes?
SSSSS [86.1K]
<span>Unsafe passes are passes with restricted line of sight, passes with cross traffic, narrow passes which are unsafe. 
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7 0
3 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
3 years ago
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