gwenspooner85

Latex is enabled, but you can't just simply copy and paste the latex here to show up. You have to press the latex button when typing a response and then enter your latex.

(a) We use the vertical asymptote, x=3, to help us solve for c. We let x be 3. We need to find c so that the denominator, x+c, is 0, and the fraction is undefined.

\(x+c=0\\3+c=0\\c=-3.\) 

We let f(x) equal to y and plug in the value of c. We then have

\(y=\frac{ax+b}{x-3}\\y(x-3)=\frac{ax+b}{x-3}(x-3)\\y(x-3)=ax+b\\xy-3y=ax+b\\xy-ax=b+3y\\x(y-a)=b+3y\\x=\frac{3y+b}{-a+y}.\) 

Now we use the horizontal asymptote y=-4 to help us solve for a. Using similar logic as we solved for c to solve for a, we get

 \(-a+y=0\\-a-4=0\\a=-4.\)

We then plug the values of a and c into \(f(x)=\frac{ax+b}{x-3}\) to get \(f(x)=\frac{-4x+b}{x-3}\). We plug in the point (1, 0) into the equation to solve for b. \(f(x)=\frac{-4x+b}{x-3}\\0=\frac{-4(1)+b}{1-3}\\0=\frac{-4+b}{-2}\\0=-4+b\\b=4.\)

Therefore \(f(x) = \frac{ax+b}{x+c}\) is \(\boxed{f(x) = \frac{-4x+4}{x-3}.}\) 

Use similar logic to solve for part (b).

Using difference of squares is the most efficient way. I don't know how to do part (b), but I can certainly explain part (a).

Something is missing from your problem.

We write the equation in 3x-7y=21 first. 

3x-7y=21

7y=3x-21

y=3/7x-3

To find the x-intercept, we plug in the value 0 for y.

0=3/7x-3

3/7x=3

x=7

To find the y-intercept, we plug in the value 0 for x.

y=3/7(0)-3

y= -3

The x-intercept is (7,0) and the y-intercept (0, -3).

To find the distance, we plug the points into the distance formula.

\(\sqrt{(7-0)^2+0-(-3)^2}\\ \sqrt{49+7}\\ \sqrt{58}\\ \)

Therefore the distance between the x and y intercept is \(\boxed{\sqrt{58}}\).

This could help: https://artofproblemsolving.com/wiki/index.php/LaTeX.

I picked up LaTeX because it was much more helpful to learn it for a math class I had to take. I learned it as classes went on, and picked up more symbols, etc. I would just say time is the main factor for learning.

Move the terms around so we have a quadratic equation.

\(2x^2-10x+35=x^2+10\\ x^2-10x-25=0\)

Now we can simply solve for x using the quadratic formula. Let a=1, b=-10, and c=-25.

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x=\frac{10\pm\sqrt{(-10)^2-4(1)(-25)}}{2(1)}\\ x=\frac{10\pm\sqrt{200}}{2}\\ x=\frac{10\pm10\sqrt{2}}{2}\\ x=5\pm5\sqrt{2} \)

So x is \(5+\sqrt{2}\ \text{and}\ 5-\sqrt{2}\).

We look at the current sum of the digits we have(not including D and E).

The sum is 7+6+7+8+9=37. So 37+D+E must be a multiple of 9. Now let's look at the multiples of 9 greater than 37:

45, 54, 63....

However, the greatest that D+E can add up to is 18, so the only possible multiples of 9 would be 45. Since there are no other possible multiples of 9, then the sum of D+E is 45-37=8.