NotThatSmart

First off, we need to put a in terms of b. This helps us identify a range. 

We get

\(a+b=1\\ a=1-b\)

Now, plugging that into the expression, we get

\(\frac{1}{a}+\frac{1}{b}= \frac{1}{1-b}+\frac{1}{b}=\frac{b+1-b}{b-b^2}=\frac{1}{b-b^2}\)

Now, we can easily find the set of all real numbers. We have

\(\{ R>1\} |\{a,b\} \subset R\) where \(1> a,b > 0\)

Thanks! :)

First, let's add up the two numbers. We get \(3578 + 2238 = 5816\)

Now, we want to convert 5816 to base 8. 

First, we have to do \(5816/8\). We get a quotient of 727 and a remainder of 0.

Next, we have \(727/8\). There's a remainder of 7 but a quotient of 90. 

Doing \(90/8\) has a quotient of 11 and a remainder of 2.

\(11/8 = 1 R 3\)

\(1/8=0R1\)

So for our final answer, we get \(13270_8\)

Thanks! :)

First, we must set some variables to complete this problem. Let's set HB as x. 

Note that \(AC^2 = AB^2 - BC^2 = (2 +x)^2 - 1^2 = x^2 + 4x + 3\)

We can write a system of equations to help us solve this question. We have

\(CH^2 = BC^2 - BH^2 = 1 - x^2 \\ CH^2 = AC^2 - AH^2 = (x^2 + 4x + 3) - 2^2 = x^2 + 4x -1 \)

Since both equations are equal to CH^2, we set both of them to equal each other, and we get

\(1-x^2 = x^2 + 4x -1 \\ 2x^2 + 4x -2 = 0 \\ x^2 + 2x - 1 = 0 \\ x^2 + 2x = 1\)

\(x^2 + 2x + 1 = 1 + 1 \\ (x + 1)^2 = 2\\ x + 1 =\sqrt{2} \\ x =\sqrt{2} - 1 \)

Now, we move on to the final step, which is to find CH. Plugging x back into the original formula, we get

\(CH^2 = BC^2 - BH^2 = 1^2 - (\sqrt{ 2} -1)^2 = 1 - (2 - 2\sqrt{ 2} + 1) = 2\sqrt {2} - 2 \\ CH = \sqrt {2\sqrt {2} - 2} = \sqrt{\sqrt {8} - 2} ≈ .91\)

So our final answer is about .91. 

Thanks! :)

We can use a really handy formula to help solve this question. 

First, we have to calculate the semiperimeter of the triangle, which is half the sum of the three side.

We get \(\frac{3+4+6}{2}=\frac{13}{2}\)

That may seem useless, but it will come in handy very soon. 

Now, we apply Heron's Formula, which states that the area of a triangle is equal to \(\sqrt{s(s-a)(s-b)(s-c)}\). Subbing everything in, We get

\(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\)

\(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)

So our final answer is \(\frac{\sqrt{455}}{4}\)

Thanks! :)

First, let's note that if AD = AB, then \(\angle ADB = \angle ABD\)

So, this means we have \(ABD = (180 - 17) / 2 = 81.5° \)

Our final answer is thus 81.5

Thanks! :)

Let's define some variables to help us solve the problem. Let's set the smallest number as n. 

We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

Since the sum of the integers add up to 28/5 times the largest number, which is n+6, we can write the equation

\(7n+21=28/5(n+6)\)

Now, we simply solve for n. We get

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

Multiplying both sides by 5/7, we get \(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

So our final answer is 9. 

Thanks! :)

We can find the diameter of the circle with the 3 x intercepts. 

The diameter of a circle is the distance between any two points that crosses the center. Since the x intercepts defintely cross the center, the diamter is

\(10-(-3)=13\)

Now, the same goes for the y intercept. Since the other y intercept must be 13 units away from 5, then we have \(5-13=-8\)

So the other y intercept is -8. 

Now, for the area, we already know the diameter, and the radius is half the diameter, so we have \(r=\frac{1}{2}13 = 6.5\)

The area of the circle is equal to \(r^2 \pi\), so plugging in 6.5, we get \(42.25\pi\) as the area. 

Thanks! :)

First, we need to simplify the equation, so let's combine all the like terms. 

We get \(-15x^2-kx+45\)

Now, in order for the solutions to be nonreal, the descriminant must be less than 0.

The descriminant is \(b^2-4ac\), so plugging in our values, we get

\((-15)^2 - 4(-k)(45) < 0 \\ 225+180k<0\\ k<-1.25\)

Therefore, the biggest interger k can be is -2. 

So -2 is our answer!

Thanks! :)