1. \(\text{We can use an integral: } \int^{2}_{-1}( x^3 + 2) dx = ? \text{ We now find the antiderivative of } x^3+2\\ \text{which turns out to be } \frac{1}{4} x^4 \text{. Then, we plug it in:} \frac{1}{4}(2)^4 - \frac{1}{4}(-1^4) = \frac{17}{4}. \)
2. \(\text{We use another integral: } \int^{3\pi/4}_{0} \sin x \: dx. \text{Take the antiderivative of }\sin x \text{, which is just }-\cos x. \text{Thus, the answer is }\\ -\cos(\frac{3\pi}{4}) - ( -\cos 0) = \frac{\sqrt{2}}{2} + 1. \)
i would assume so
i don't think i made a mistake there.... check the graph again....
you can use a calculator to check, you know
ok, no need to say it's due tmrw, cuz this is an aops thing
the solution:
Let's say that the radius is equal to r. Let's say the midpoint of CD is P. Then, we know that PC= 9. Draw OP, and notice that OP is perpendicular to CD. Also, since \(\angle ABD \) = 90, that means that AOPB is a rectangle, which implies AB=OP= 12. Using the pythagorean theorem, the radius is equal to \(\sqrt{9^2 + 12^2}, \text{which is equal to } \boxed{15}\). Just use pi r ^2 to find the area, which is 225 pi
7. arcsin 7/15 = 27.818
8. arccos 8/25 = 57.769
a. when x = (-6, -3) and (3,13)
b. when x = (\(-\infty\), -6) and (-3,3)
c. when x= (0,4) and (-8, -4)
d. when x = (4,13) and (-4,0)
e. when x =(\(-\infty\),-8)
f.the domain is when x = (\(-\infty \),13) and the range is when y = (-9,7)
1. Undefined, as 0^0 is still being debated. Otherwise, it would be geometric, because it is the sequence 0^0, 1^1, 2^2, 3^3
2. This is an arithmetic sequence because each term decreases by 1.5.
3. Neither, as this can be recursively defined as something with even/odd cases.
5. I think neither, as it each term decreases by n+1
nonzero digits are significant, so it would be 20cm and 506.7cm
a. The positive parts are -6
b. The negative parts are -3
c.It is increasing at -8
d. decreasing at 4
e. The constant part is x<-6
f.The domain of the function is x<=13. The range of the function is -9<=y<=7.
