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mezya [45]
3 years ago
5

PLEASE HELP

Mathematics
1 answer:
zubka84 [21]3 years ago
6 0
If there are 1.5 times as many pairs of walking shoes as there are running shoes on display, then it is best to take each amount of running shoes and multiply it by 1.5, if doing that equals the amount of walking shoes, then it would be correct. Which in this case, would be D.
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What is the product of the polynomials below?<br> (7x2+2x+4)(2x+5)
Gelneren [198K]

(16x+4)(2x+5)

Step-by-step explanation:

4 0
3 years ago
Solve for x: 2|х - 3 |+1 = 7
Rasek [7]
When using problems like this use mathaway love
5 0
3 years ago
Classify the following polynomials by the number of terms and degrees<br> -2x^3 +5x
cestrela7 [59]

Answer:

A third degree binomial

Step-by-step explanation:

The highest power is 3 hence the 3rd degree

and there are 2 terms hence the bi-nomial

8 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
Expand and simplify 9a+3(8-2a).
olchik [2.2K]
Hi there

9a + 3(8-2a)
9a + 24 - 6a
(9a -6a) +24
3x + 24


I hope that's help ! 

3 0
3 years ago
Read 2 more answers
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