Solve 4 brainliest xx

Find the point, M, that divides segment AB into a ratio of 5:5 if A is at (0, 15) and B is at (20, 0).

Airida [17]

C

a ratio of 5 : 5 simplifies to 1 : 1, which basically means we require the midpoint of the line segment

using the midpoint formula

M = [ $\frac{1}{2}$ (0 + 20 ), $\frac{1}{2}$ (15 + 0)] = (10, 7.5 )

5 0

2 years ago

Read 2 more answers

What is the missing number in the table?

mariarad [96]

Answer:

1296

Step-by-step explanation:

the y is multiplying by six each time 6*6 = 36

36*6 = 216

216*6 = 1296

2196 = 7776

4 0

2 years ago

How much money should be deposited today in an account that earns 6% compounded monthly so that it will accumulate to $1000000 (

aleksandr82 [10.1K]

Answer:

The principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.

Step-by-step explanation:

Given

Accrued Amount A = $1000000

Interest rate r = 6% = 0.06

Time period t = 45 years

Compounded monthly n = 12

To determine:

Principle amount P = ?

Using the formula

$A\:=\:P\left(1\:+\:\frac{r}{n}\right)^{nt}$

$P\:=\frac{A}{\left(1\:+\:\frac{r}{n}\right)^{nt}}$

substituting A = 1000000, r = 0.06, t = 45, and n = 12

$P\:=\frac{1000000}{\left(1\:+\:\frac{0.06}{12}\right)^{12\cdot 45}}\:$

$=\frac{1000000}{1.005^{540}}$

$P = 67659.17$ $

Therefore, the principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.

7 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: