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Dmitriy789 [7]
3 years ago
9

I need help. !!! Please explain this to me

Mathematics
1 answer:
Airida [17]3 years ago
5 0
It seems appropriate to use the linear regression function of your calculator for this. Looking at the data, you know the slope is generally positive, so only the 1st and 3rd choices are potentially viable.

A calculator shows the correlation coefficient (r) rounds to 0.816, so the appropriate selection is ...
  A. 0.816

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45 tens is the same as
MakcuM [25]
Tens refers to the unit 10,
This means that 45 10s is equal to:
45*10 = 450

Hope this helps! :)
8 0
3 years ago
Read 2 more answers
A number line is shown. Which expression represents the distance between two points, X and Y, on the number line?
IrinaK [193]

Answer: C. |-4 - 3|

<u>Step-by-step explanation:</u>

x = -4

y = 3

The distance between them is the difference. The mathematical operation for difference is subtraction.

Distance is a unit of measurement. Measurements cannot be negative so the absolute value sign is used to ensure the difference is positive.

4 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
9x (-2y + 5x) - 4 xy (6x-3y) please the person who answers will be marked as brainliest​. please show working​
Alenkasestr [34]

Answer:

See explanation

Step-by-step explanation:

9x(-2y+5x)-4xy(6x-3y)

By the distributive property this is equal to:

9x(-2y)+9x(5x)-4xy(6x)-4xy(-3y)=\\\\-18xy+45x^2-24x^2y+12xy^2

Hope this helps!

4 0
3 years ago
If 3x + 11 = 4x - 5, what is x?
Andrew [12]

Answer:

16v

Step-by-step explanation:

3x + 11 = 4x - 5

11+5 = 4x -3x

x = 16

I hope im right!!

4 0
2 years ago
Read 2 more answers
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