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Harlamova29_29 [7]
3 years ago
9

12 workers build a wall in 10hours. How long do 5 workers build a wall​

Mathematics
2 answers:
snow_lady [41]3 years ago
5 0

Answer:

24 hours

Step-by-step explanation:

If 12 workers take 10 hours to build a wall, assuming they build at the same rate, it would take one worker 120 hours to build a wall.

By this logic, it would take 120/5 or 24 hours for 5 workers to build a wall

andre [41]3 years ago
5 0

5 workers will build a wall in 24 hours.

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The answer is 23x - 16
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Write the equation in slope intercept form with a slope of 3/4 and passes through (-4,-5)
attashe74 [19]

Answer:Use the slope −2 and the point (−4,−5) to find the y-intercept.

y=mx+b

⇒−5=(−2×−4)+b

⇒−5=8+b

⇒b=−13

Write the equation in slope intercept form as:

y=mx+b

⇒y=(−2)x−13

Hence, the equation of the line is y=−2x−13.

Step-by-step explanation:

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And the second one is dividing.

Step-by-step explanation:

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3 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

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Which graph represents the polar curve r = 4cos(30)?
stepan [7]

The equation of the polar curve r = 4\cos(3\theta) is an even cosine function.

<h3>Polar curves</h3>

In geometry, polar curves are curves that are represented by the polar coordinates such that they are drawn around a fixed point

The equation of the polar curve is given as:

r = 4\cos(3\theta)

To graph the polar curve, we make use of the following representations:

  • r represents the y-axis
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Using the above representation, we can graph the polar curve

See attachment for the graph that represents r = 4\cos(3\theta)

Read more about polar curves at:

brainly.com/question/4618020

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2 years ago
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