TheXSquaredFactor

The gradient (usually referred to as the slope) of a line indicates its steepness. There is an equation that allows one to calculate the gradient. It is the following:

\(m=\frac{y_2-y_1}{x_2-x_1}\)

m = slope of line

Before we can understand this equation, let's try putting it to use on some lines:

I am assuming that you want this equation solved using the quadratic formula. I would use that method, too. Of course, let's remind ourselves of the quadratic formula:

In a quadratic function in the form \(ax^2+bx+c=0\).
 

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

The question, however, asks for how much time elapsed when it (I assume a thrown ball) reached a height of 3 meters

Now that the equation is in the form of \(ax^2+bx+c=0\), let's solve for how much time, t, it took for it to reach 3 meters:

The question asks for the first time that the ball hits 3 meters. Of course, we do not know the exact value of either of our solutions currently. We can utilize logical thinking to figure it out, however. 

If we pretend as if \(\sqrt{30}=5\), as that is an OK approximation, we can simplify both solutions into 

\(t=\frac{4}{5}-\frac{1}{2}\) and \(t=\frac{4}{5}+\frac{1}{2}\)

We can infer that \(\frac{4}{5}-\frac{1}{2}\) is still positive but closer to 0 because 4/5 is greater than 1/2. Subtracting the two would make it closer to 0. Adding both would make the numbers further from 0.

Therefore, the first time it goes to the height of 3 meters is after \(\frac{4}{5}-\frac{\sqrt{30}}{10}\) that much time elapsed. \(\frac{4}{5}-\frac{\sqrt{30}}{10}\approx0.252277\). 

Although I am not a financial expert, I believe that I have somehow heard about such advice before, so I would say that #3 is true.

There are 2 things that must be true in order to use the absorption law

1. Operators must be different

2. Term outside parentheses must also be inside the parentheses, too

Is that the case here? Yes!

Therefore, your expression simplifies to P.

If we think about the expansion of a binomial of a big power (like 7). The expansion looks like the following:
 

\((a+b)^n={n\choose 0}a^nb^0+{n\choose 1}a^{n-1}b^1+...+{n\choose n}a^0b^n\)

Knowing that this pattern exists, let's utilize that in the current expansion of the expression:

\((2x+3)^7={7\choose0}(2x)^7(3)^0+{7\choose1}(2x)^6(3)^1+{7\choose2}(2x)^5(3)^2+{7\choose3}(2x)^4(3)^3+{7\choose4}(2x)^3(3)^4+...\)

Why don't I care about the other terms? Well, I only care about the terms that has the x^5. If it doesn't have that, then I can ignore that term. In this case the appropriate term is \({7\choose 2}(2x)^5(3)^2\). Let's evaluate it:

The formula for the choose function is the following:
 

\({x \choose y}=\frac{x!}{y!(x-y)!}\)

Let's apply that to \({7\choose 2}\) first:

Ok, now let's continue with our problem:
 

Therefore, the coefficient in front of the x^5 is 6048.

Greetings to you, too :)

If your calculator evaluates the expression \(6/2(2+1)\) as 1, you need to use a different calculator. 

To say that 2 nonlinear lines intersects at a common point is correct. However, it is not given info that they are indeed nonlinear; you must prove it.

To do this problem, evaluate n at the values from the positive integers n=3 to n=7

When n=3:

Now, let's evaluate it for n=4:

n=5:

n=6:

Now, I do not have to evaluate 7 as I can infer, based on the information above, that the next value in the series will be the sum of the last number in the series and 2. Therefore, 2n+3 when n=7, the expression evaluates to 17.

Now, our last step is too add the series together: