What is 3 divided by .075

2(3) squared answer and steps.

olga55 [171]

Step-by-step explanation:

2(3)

2×3=6

6×6=36

36 is the correct answer i hope it helps you

4 0

2 years ago

You deposit $9,800 in a money-market account that pays an annual interest rate of 5.8%. The interest is compounded quarterly. Ho

Alona [7]

<span>$16,620.80

Hope this helped!</span>

8 0

2 years ago

Read 2 more answers

i need help please! I think I know the answer but for some reason my brain's telling me that it's wrong so if someone could expl

Ivenika [448]

Considering the slopes of the segments, the correct option is:

D. No, because the triangle ABC doesn't have a pair of perpendicular sides.

<h3>When are lines parallel, perpendicular or neither?</h3>

The slope, given by <u>change in y divided by change in x</u>, determines if the lines are parallel, perpendicular, or neither, as follows:

If they are equal, the lines are parallel.

If their multiplication is of -1, they are perpendicular.

Otherwise, they are neither.

Here, we have to find if there are perpendicular segments, that is, if two slopes multiplied have a value of -1, then:

mAB = (-7 - 9)/(11 - 1) = -8/5.

mAC = (3 - 9)/(-9 - 1) = 3/5.

mBC = (3 - (-7))/(-9 - 11) = -1/2.

No sides are perpendicular, hence option D is correct.

More can be learned about slopes at brainly.com/question/20847660

#SPJ1

6 0

1 year ago

What is the volume of a sphere with a radius of 5.9 in, rounded to the nearest tenth of

Olegator [25]

Answer:

6

Step-by-step explanation:

If u round it to the nearest 1

6 0

2 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

2 years ago