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Area of triangle ACE = 150 square units

To give a little hint, the formula of interest rate to get the money you have is: \(A = P(1 + {r\over n})^{nt}\) where:

A = Amount you have after a certain period of time.

P = Original amount, principal amount.

r = Annual interest rate.

n = Number of times the interest compounds each year.

t = time in years.

Hope that helps and gives you a little lift on how to get started with this problem :)

I think applying the formula is the smartest way to do it, I did it a little unorthodoxly, thank you Guest! 

The answer would be D.

The reason so can be determined by taking a look at each option.

Option A would get us \(x^4 = m^2\), and wouldn't help us in anyway.

Option B would get us \({x^2\over m}=1\), and would only get us what the value of \(x^2\over m\), but that isn't what we are looking for.

Option C would get us \({x^2\over2}={m\over2}\), and would get us nowhere.

Option D would get us \(x = \sqrt{m}\), and get us the value of \(x\) in terms of \(m\), which is more helpful, and since \(x\) is normally the unknown or what we are trying to solve,

Option D is the best way to solve the equation \(x^2 = m\).

Use the formula for the sum of an arithmetic sequence:

S==96/2 * [2* 20  +  (3*95)], solve for S

S ==48 * [40  +  285]

S ==48 * 325 ==15,600

Actually \(({4\over3})^3\) is \(4^3\over3^3\), that simplifies to \(64\over27\).

Thus the correct answer to this problem is \(64/27\). 

The 5th term is 20. The pattern is the n-th term is 3(n-1) + 8, so the 100-th term of the pattern is 3(99) + 8 or 305. Thus, all the numbers added from 20 to 305 would be added together, with a difference of 3 between each number. Since we know there are a total of 96 numbers added together, we can subtract 17 from each number, then divide each number by 3. Then the numbers from 1 - 96 would be added together, which equals 4656.

Then we multiply 4656 by 3 to get 13968. Then since we subtracted 17 from 96 numbers, we have to add (17 x 96) to 13968. That gets 15600. Thus, the answer is 15600.

Correct me if I got it wrong :D

thank you for your solution!

When the ball hits the ground, y=0

0 = -4.9 t^2-3.5t+12.4      Use Quadratic formula to solve for 't'       Throw out the negative value  

'

The question doesn't ask for it,

but the two positive numbers are 3 and 8.  

f(x) is a polynomial of degree 4 or greater such that f(1) = 2, f(2) = 3, and f(3) = -4.

Find the remainder when f(x) is divided by (x - 1)(x - 2)(x - 3).

f(x) = P(x)*(x - 1)(x - 2)(x - 3) + R(x) where P(x) is some polynomial and R(x) is the remainder ax² + bx + c

f(x) = P(x)*(x - 1)(x - 2)(x - 3) + (ax² + bx + c)

f(1) = a + b + c = 2

f(2) = 4a + 2b + c = 3

f(3) = 9a +3b + c = -4

Solving for a, b , c => a = -4, b = 13, c = -7

R(x) = -4x² + 13x - 7. 

AEFGD = 91.94

There are \(200\) integers that are divisible by \(2\) 

There are \(133\) integers that are divisible by \(3\)

We need to account for multiples of both \(2\) and \(3\)

There are \(66\) intergers that are divisble by \(6\). 

So, there are \(400-200-133+66 = \color{brown}\boxed{133}\)

i might have messed up somewhere, these questions are easy to slip up on.

Let g(x) =  p(x)*(x²-x-6) + R(x)   where p(x) is some other polynomial and R(x)= x + 5.

When x = 3   g(3) = p(3)*(3^2-3-6) + (3+5)   ie  x^2-x-6 = 0, so all that is left of g(3) is 3 + 5.  i.e. g(3) = 8.

Nevermind wrong thing!!!

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ShockFish

\(\color{brown}\boxed{-7\over9}\)

The equation can be written as \(6(j^2-j-2)\). 

We can divide the equation by \(6\) to get \(j^2-j-2\). 

This is: \(j^2-j-2=0\)

Add \(2\) to both sides: \(j^2-j=2\)

Rewrite: \((j-0.5)^2 = 2+0.25\)

So: \(6(j-0.5)^2 = 6j^2-6j+1.5\)

We need to subtract \(13\) to get the equation, so we have: \(6(j-0.5)^2-13\)

Thus, the answer is \({-13\over-0.5} = \color{brown}\boxed{26}\)

The distance between \(A\) and \(B\) is \(12\) across, and \(6 \) up.

Because the distance between \(B\) and \(C\) is half of the distance, \(C\) will be \(6\) units across, and \(3\) units above \(B\), meaning that \(C\) is \(\color{brown}\boxed {(20,7)}\)

I brute forced it. 

\(A = 15, B = 4, C = 10, D = 6\)

So, \(\color{brown}\boxed{A=15}\)

\({4\over 3}^3 = {4\over 3}\times {4\over 3}\times {4\over 3} = {64\over 9} = \color{brown}\boxed{7 {1\over9}}\)

The smallest n that works is 32.

The maximum value is 6*sqrt(2).

The area of AEFGD is 25.

Discriminant of Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

b^2 - 4ac  >=0

15^2 - 4(2)(b) >= 0

225 >= 8b

28.125 >= b                 sum = 28 + 27 + 26.......1 = 406    for positive integers