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Step2247 [10]
3 years ago
8

paul says that the area of a triangle with base 16 meters and height 5 meters is 80 square meters paula says that the area is 40

square meters who is correct
Mathematics
1 answer:
rosijanka [135]3 years ago
3 0

Paul forgot to divide by 2. The area of a triangle is A = 1/2 * b * h

Paul just multiplied the Base and height together. 5*16 = 80.

Paula used the formula correctly. The answer is 40 square meters.

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3 years ago
11111111111111111111111
Nat2105 [25]

Answer:

1) Zero based on (-16·t - 2) is t = -1/8 second

2) Zero based on (t - 1) is t = 1 second

Step-by-step explanation:

The given functions representing the height of the beach ball the child throws as a function of time are;

y = (-16·t - 2)·(t - 1) and y = -16·t² + 14·t + 2

We note that (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

Therefore, the function representing the height of the beachball, 'y', is y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The zeros of a function are the values of the variables, 'x', of the function that makes the value of the function, f(x), equal to zero

In the function of the question, we have;

y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The above equation can be written as follows;

y = (-16·t - 2) × (t - 1)

Therefore, 'y' equals zero when either (-16·t - 2) = 0 or (t - 1) = 0

1) The zero based on (-16·t - 2) = 0, is given as follows;

(-16·t - 2) = 0

∴ t = 2/(-16) = -1/8

t = -1/8 second

The zero based on (-16·t - 2) is t = -1/8 second

2) The zero based on (t - 1) = 0, is given as follows;

(t - 1) = 0

∴ t = 1 second

The zero based on (t - 1) is t = 1 second

4 0
3 years ago
1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle
zimovet [89]

Answer: \left |  2t-5\right |,\ 2,\ 2t-5

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

s(t)=t^2-5t+1

Speed of the particle is given by

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left |  2t-5\right |

Acceleration of the particle is

\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2

velocity can be negative, but speed cannot

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5

3 0
3 years ago
What is the circumference of this circle given a radius of 7 in.?
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7 0
3 years ago
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Echa t-shirt is 6 dollars and ecah hat is 5
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