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.
Find a common denominator which is 12
1*3=3 4*3=12 so 3 3/12
then 5*2=10 6*2=12 so 2 10/12
2 10/12+3 3/12= 5 13/12
13/12= 1 1/12 plus 5= 6 1/12
so your answer would be 6 1/12
Answer:
x = 3
Step-by-step explanation:
2x + 8 = 14
2x = 6
x = 3
<h2>
Answer:</h2><h3>
A. Domain </h3>
The domain of a function is the x-values that the graph applies to. This means that the domain is whatever x-values the graph crosses. All vertical parabolas (like the one pictured) have a domain of all reals. This is because any x-value could be plugged into the function and provide a y-value. while it may not seem like it, that graph will cover every single x-value in existence.
<h3>
B. Range</h3>
The range is similar to the domain but is for y-values. So, the range is whatever y-values the graph applies to and crosses. As you can see from the graph, there are no y-values above 1. This means the range is y≤1.
<h3>
C. Increasing Interval</h3>
A graph is increasing when the y-values are increasing. So, on the parent function of a parabola, the graph increases to the right and decreases to the left. However, this graph is inverted and shifted to the left, so the interval will also be flipped and shifted. In this case, the graph increases from -∞ to 2.
- Increasing Interval = [-∞, 2]
<h3>
D. Decreasing Interval</h3>
The decreasing interval is very similar to the increasing interval. This interval applies when the y-values are decreasing as the x-values increase. For a parabola, the increasing and decreasing intervals always meet at the x-value of the vertex, which is 2 on this graph. The y-values decrease during the interval 2 to ∞.
- Decreasing Interval = [2, ∞]
<h3>
E. Opening</h3>
The direction of a parabola is decided by the sign (+ or -) of the leading coefficient. Positive coefficients open up and negative opens down. As you can see from the graph, the sides of the parabola point downwards. This means that the leading coefficient must be negative.
<h3>
F. Min and Max</h3>
A parabola will always only have a min or a max, never both. If a graph opens up it has a min because there is one y-value which is the minimum possible y-value. Graphs that open downwards have a maximum because there is one y-value that is the largest possible. So, this graph has a maximum of 1 because that is the largest possible y-value.
Answer: divide the big number by 2
Step-by-step explanation: dude it literally tells you the answer almost xD jsut divide the big number by 2
