Plz help PRE CALC !!!!!!!

I just need one more answer to this. Please someone help!

den301095 [7]

Answer:

(1,0)

(-2,3)

(5,24)

Step-by-step explanation:

To solve this you can can just plug in the x and y values and see which work

y = x²-1

Lets test (0,1):

1 = 0²-1

1 = -1

This pair does not work, because 1 does not equal -1

Lets test (1,0):

0 = 1²-1

0 = 0

This does work, because 0 equals 0

Lets test (3,5):

5 = 3²-1

5 = 9 - 1

5 = 8

This pair does not work, because 5 does not equal 8

Lets test (5,24):

24 = 5²-1

24 = 25 -1

24 = 24

This pair does work, because 24 equals 24

Lets test (-2,3):

3 = (-2)²-1

3 = 4-1

3 = 3

This pair does work, because 3 equals 3

Lets test (-4,-17):

-17 = (-4)²-1

-17 = 16 - 1

-17 = 15

This pair does not work, because -17 does not equal 15

So the pairs that are on the graph are:

(1,0)

(-2,3)

(5,24)

4 0

3 years ago

You don't have to answer them all but help me out !!

Mandarinka [93]

Answer:

1. 10n

2. 3x^2z^2

3. 2(9n^3 - 3n^2 + 3n - 1)

4. (Put the same thing)

5. There are no common factors so it is completely factored out.

Step-by-step explanation:

For 1, you gotta find the gcf of the numbers and the variables. So, the common factor of 90, 70, 70, and -80 is 10. Nothing else greater than 10 divides these numbers. For the variables, only "n" is the GCF, as there is one "n" and all the other numbers have n^2 and etc.

For 2, same thing, you have to find the gcf for the numbers and the variables.

3. same thing ;-; (gcf is 2 btw, nothing divides the variables because 2 has no "n".

4. theres nothing that divides the numbers or the variables. So, the gcf is just 1.

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: