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loris [4]
3 years ago
6

If k, m, and t are positive integers and k + m = t , do t and 12 have a common factor 6 4 12 greater than 1 ?

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
7 0
You asked this question a week ago im pretty sure tht ur not coming back to see this answer
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Musya8 [376]
The answer to your question is 10/3
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What is the slope of the line that passes through points (-1,5) and (2,-3)
Sonbull [250]

Answer:

-8

Step-by-step explanation:

3 0
3 years ago
Cindy and Zoey work at a store. Both girls earn $6.25 per hour
wolverine [178]

Answer:

Step-by-step explanation:

I think this is your full question, right?

<em>Cindy and Zoey work at a store. Both girls earn $6.25 per hour. During a normal week Cindy works 15 hours and  Zoey works 20 hours. The expression 6.25(15) + 6.25(20) can be used to calculate the total amount of money that the  girls earned in one week. Which expression shows another way to calculate the amount of money the girls earn in  one week?  </em>

My answer:

The expression shows another way must be equal to: <em>6.25(15) + 6.25(20)  </em>

<=> 6.25 (15+20)

Because both girls earn the same per hour.

8 0
3 years ago
QUICK HELP!!!!!!!!!!!!!!!!
JulijaS [17]

Answer:

Its C im pretty sure. Also loveee your pfp!

3 0
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

5 0
3 years ago
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