Answer:
Option (1)
Step-by-step explanation:
Lines X and Y are intersecting at a point.
Two pairs of vertical angles have been formed at the point of intersection.
1). ∠1 and ∠3
2). ∠2 and ∠4
It's given that m(∠1) = m∠3 = 55° [Since pair of vertical angles are equal in measure]
If line Y is rotated 5° clockwise about the point of intersection of both the lines,
∠1 and ∠3 both will increase by 5°.
Therefore, measure of angles 1 and angle 3 will become (55° + 5° = 60°).
Option (1) will be the answer.
Answer:
the answer is b
Step-by-step explanation:
i took the test it is b. The sum of the areas of faces A, B, and C.
Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
Both integrals are quite immediate: you only need to use the power rule
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
Answer:
In words the answer is between t=0 and t=2.
In interval notation the answer is (0,2)
In inequality notation the answer is 0<t<2
Big note: You should make sure the function I use what you meant.
Step-by-step explanation:
I hope the function is h(t)=-16t^2+32t because that is how I'm going to interpret it.
So if we can find when the ball is on the ground or has hit the ground (this is when h=0) then we can find when it is in the air which is between those 2 numbers.
0=-16t^2+32t
0=-16t(t-2)
So at t=0 and t=2
So the ball is in the air between t=0 and t=2
Interval notation (0,2)
Inequality notation 0<t<2
