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Paladinen [302]
3 years ago
7

Determine the value of b that allows each expression to be factored. (X^2-bx-8)

Mathematics
1 answer:
Kipish [7]3 years ago
7 0
The answer is b=2
cuz (x-4)(x+2) multiplies
into x^2-2x-8
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Two gears are both connected to the same belt. The smaller of the two gears has a radius of 4ft & the larger of the two gear
krek1111 [17]

Answer:

\omega_1 = \frac{5Rad}{3s} and \omega_2 = \frac{35Rad}{12s}

Step-by-step explanation:

Given that the larger gear is turning at the rate 50rpm. The angular velocity, \omega_1, of the larger gear can be calculated as follows:

\omega_1 = \frac{50rev}{m} to convert that to secs, we know that 1 rev = 2πRad and 60secs make a min, so we have:

\omega_1 = \frac{50rev}{m} \times \frac{2\pi Rad}{1rev} \times \frac{1m}{60s}

That gives

\omega_1 = \frac{100Rad}{60s}

Which gives:

\omega_1 = \frac{5Rad}{3s}

This is the angular velocity of the larger gear in rads/sec.

In order to get the angular velocity of the smaller gear, we know from the question that both gear are connected to the same belt. This implies that they have the same linear velocity. Thus let us find the linear velocity of the larger gear. This is given by

v = r_1 \times \omega_1 where v stands for linear velocity and r_1 for radius of larger velocity.

We are given the radius of the larger gear as 7ft so we have

v = 7ft \times \frac{5Rad}{3s} = \frac{35ft}{3s}

To get the angular velocity of the smaller gear, we use

v = r_2 \times \omega_2 where r_2 is the radius of the smaller gear which is given as 4.

Since linear velocity is the same, we have

\omega_2 = \frac{35ft}{4 \times 3s}

\omega_2 = \frac{35Rad}{12s}

5 0
3 years ago
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