gwenspooner85

First, we note that r must be positive, since otherwise \(\lfloor r \rfloor + r\) is nonpositive. Next, because \(\lfloor r \rfloor + r\) is an integer and \( \lfloor r \rfloor + r=12.2\), the decimal part of r must be 0.5. Therefore, r=n+0.5 for some integer n, so that \(\lfloor r \rfloor + r\) and \(\lfloor r \rfloor + r = 2n+0.5 =16.5\). Therefore, n=8, and the only value of r that satisfies the equation is \(\boxed{r=8.5}\).

Oh never mind I see your messages. I responded to them.

1. The discriminant of a quadratic is b^2-4ac.

In the equation 5x^2-2x+8, a=5, b= -2, c=8. So plugging in the values of the variables, b^2-4ac=(-2)^2-4(5)(8)=4-160= -156.

2.

 \(\frac{x+4}{x+5}=\frac{x-5}{2x}\\ 2x^2+8x=x^2-25\\ x^2+8x+25=0\)

Now we just have a quadratic equation and we solve to get  \(x=-4+3i, -4-3i\).

You're welcome.

We find the value of f(x) that results in 3(f(x)=0). 

For 3-x, x=3. For -x^3+2x^2+3x, x=0, 3, -1. None of these satisfies x>3, so f(0) cannot go into the second piece of the function. It can only go in the first, so f^-1(0) is 3.

Then we find the value of f(x) that results in 6. 

For 3-x, x=-3. For -x^3+2x^2+3x, x=2, \(\sqrt{3}\ \text{or}-\sqrt{3}\). None of these satisfies x>3, so again, f(6) cannot go into the second piece of the function. It can only go in the first, so f^-1(6) is -3.

Adding these together, 3+(-3)=0.

Help you with what?

1. 270 counterclockwise: (x,y)-->(y,-x)

So (3,-4)--> a.(-3,-4)

2. R is 4 units below the x axis. When reflected, it is 4 untis above the x axis. So R' would be a. (3, 4)

3. Translating 13 units up is essentially the same as adding 13 to the y point so -4+13=9. So R' would be d. (3,9).

4. A scale factor of 1/5 results in 1/5 of all the points.

L(0,10)-->L'(0,2)

M(-5, -5)-->M'(-1,-1)

J(0,-10)-->J'(0,-2)

K(5,-5)-->K'(1,-1)

So the answer is b.

For f(2)=5, 2 is greater than 0 so it's substituted into the equation ax+3 to get 2a+3=5.

For f(0)=5, 0 is directly equal to 0 so it's substituted into the equation ab to get ab=5.

For f(-2)=-10, -10 is less than 0 so it's substituted into the equation bx+c to get -2b+c=-10.

We have a system of equations now:

2a+3=5

ab=5

-2b+c=-10

From the first equation, 2a+3=5, a=1.

We substitute the value of a into ab=5, and get b=5.

Finally, from the last equation, -2b+c=-10, c=0.

So a+b+c=6.

We are given the expressions of each angle and we know the angles in a quadrilateral add up to 360.

So we form an equation:

q+5q+15+3q-25+4q-20=360

13q-30=360

13q=390

q=30

The largest angle is 5q+15, which is 5(30)+15=\(165^\circ\).

\(x-1/x=3\\ x^2-1=3x\\ x^2-3x-1=0\\ x=\frac{3\pm\sqrt{13}}{2}\\ \)

So

x+1/x

=\(\frac{3+\sqrt{13}}{2}+\frac{1}{3+\frac{\sqrt{13}}{{2}}}\\ =\sqrt{13}\)

x+1\x

\(\frac{3-\sqrt{13}}{2}+\frac{1}{3-\frac{\sqrt{13}}{{2}}}\\ =-\sqrt{13}\)

So I believe the possible values of x+1/x are \(\boxed{\sqrt{13}\ \text{and}\ -\sqrt{13}.}\)