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3 years ago

9

Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 6

0 seconds. If they start rotating now from the same point, when will they be at the same starting point again?

1 answer:

lara [203]3 years ago

5 0

Find the rate in which both balls make one full rotation around the track.

Ball A:

$120 / 4 = 30$

It takes 30 seconds for Ball A to make a full rotation.

Ball B:

$60 / 3 = 20$

It takes 20 seconds for Ball B to make a full rotation.

Find the least common multiple of 20 and 30 to find the time at which they'll meet.

Multiples of 20: {20,40,60}
Multiples of 30: {30,60}

60 is the least common multiple of both 20 and 30.

The balls will meet again at 60 seconds.

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PLEASE HELP ME!!

Ludmilka [50]

Answer:

<h2>The population will reach 1200 after about 2.8 years</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

The population of a certain species of bird in a region after t years can be modeled by the function P(t) = 1620/ 1+1.15e-0.42t , where t ≥ 0. When will the population reach 1,200?

According to question we are to calculate the time t that the population P(t) will reach 1200.To do this we will substitute P(t) = 1,200 into the equation and calculate for the time 't'.

Given;

$P(t) = \frac{1620}{1+1.15e^{-0.42t} } \\\\at \ P(t)= 1200;\\\\1200 = \frac{1620}{1+1.15e^{-0.42t} }\\\\cross\ multiplying\\\\1+1.15e^{-0.42t} = \frac{1620}{1200} \\\\1+1.15e^{-0.42t} = 1.35\\\\1.15e^{-0.42t} = 1.35-1\\\\e^{-0.42t} = \frac{0.35}{1.15}\\ \\e^{-0.42t} = 0.3043\\\\Taking \ ln\ of\ both\ sides\\\\lne^{-0.42t} = ln0.3043\\\\-0.42t = -1.1897\\\\t = \frac{-1.1897}{-0.42} \\\\t = 2.8 years\\\\$

The population will reach 1200 after about 2.8 years

6 0

3 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

so

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

4 years ago