ilorty

Unless my counting skills are plummiting, the shape created by the 6 lines should form a hexagon. And, in a hexagon, there should be a total of 720 degrees. (angle sum). So, all we need to do is find the angles that we do know, and solve for x. We know that:

BCD=45

CDE (interior)=360-120=240

DEF=45

AFE=180-60=120

BAF=180-60=120

Now that we have all the angles that we know, it is time to solve for x. So, 720-(45+45+240+120+120)=720-57=150.

Therefore, x=150.

:)

To answer the question, we need to see which way the parabola is facing. Since it is facing up, the smallest point is the vertex. (if it faced down, then there would be no smallest point) Using the vertex formula (this helps us find at what value of t we get the vertex), we see that -b/2a=8/2=4. 

Therefore, the smallest value of t is 4. 

First step is find the area of the base. We see that it is 5 times 5, which is 15. Now, we have to find the combined area of the 4 triangles. This is 2 times 6 times 5, which is 60. Add 25 and 40, and we get 85. Therefore, the answer is a.

:)

Im not sure if there is an equation.....

Anyways, we can first find the area of the square. That is 0.5^2, which is 0.25. The area of the triangles are 0.5(0.4)(2), which is 0.4. Add those together, and you get 0.65.

EDIT: THanks guest, i need to practice my addtion, lol

:)

To do this, let's first expand. (i couldn't find an easier way...)

So, (u+4)(u-1)-(u-3)(u+6) is u^2+3u-4-(u^2+3u-18), which is 14.

PS After some thought about the answer I got, I think there might be an easier way....but i dont see it yet....

:)

First, let's find the area of the circle. We see that the radius is 4. Plug that into the area of a circle equation, and you get 16pi. Now, we need to find the area of the triangle. To do this, we see that QO=OP, and therefore OP=4. So, we use the area of a triangle equation, and we get that the area is 8. Subtract that from 16pi, and the answer is 16pi-8. 

Sorry, my LaTeX did not render properly....again.....

Is this the number you want: \(0.\overline{481}\)? This question is actually really simple.

All repeating numbers are going to have a denomintator with repeating 9s for the amount of digits there are after the decimal point. This means, that \(0.\overline 3\) is 3/9,or 1/3, and \(0.\overline {49}\) is 49/99, and so on. Therefore, 0.481 repeating is 481/999, which is simplified to 13/27.

:)

Domain:

Because there is a square root, we see that x^2 can be no greater than 9. (or else there will be an imaginary number, which is not allowed in functions) So, we can "unsquare" (not square root, there is a difference) 9 and get x=-3 and 3. Therefore, the domain is -3<=x<=3, which is [-3, 3] in interval notation.

Range:

We can take the domain to help us find the range. We see that the greatest number x^2 can be is 9, so we solve for sqrt(9-9)+4. That is 4. We see that the smallest number x^2 can be is 0. Plug that in, and we get 7. Therefore, the range is [4, 7].

:)