Answer:
x=1
Step-by-step explanation:
3x+8x-8=3
11x=3+8
11x=11
x=11/11
x=1
Answer:
Option (2)
Step-by-step explanation:
Option (1)
In the interval [-1, 1] Or -1 ≤ x ≤ 1, value of the function will be represented by the y-values.
Since, the graph is below x-axis in the interval -1 ≤ x ≤ 0.5, function will be negative.
And in the interval 0.5 ≤ x ≤ 1, graph is above the x-axis, function will be positive.
Option (2)
In this option, graph is below the x-axis in the interval [-1, 1].
Therefore, the given graph is negative in this interval.
Option (3)
In this graph, function is negative and positive both in the interval -1 ≤ x ≤ 1.
Option (4)
In this graph function is completely above the x-axis in the interval -1 ≤ x ≤ 1.
So this function is positive in the interval [-1, 1].
Therefore, Option (2) will be the answer.
Here's the equation for one year:
0.03(8000) + 8000
For four years:
4(0.03(8000)) + 8000
0.12(8000) + 8000
To make it simpler/smaller:
1.12(8000) = 8960
You can make 960 more dollars, and you now have $8960
Step-by-step explanation:
(a)
Using the definition given from the problem
![f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\](https://tex.z-dn.net/?f=f%28A%29%20%3D%20%5C%7Bx%5E2%20%20%5C%2C%20%3A%20%5C%2C%20x%20%5Cin%20%5B0%2C2%5D%5C%7D%20%3D%20%5B0%2C4%5D%5C%5Cf%28B%29%20%3D%20%5C%7Bx%5E2%20%20%5C%2C%20%3A%20%5C%2C%20x%20%5Cin%20%5B1%2C4%5D%5C%7D%20%3D%20%5B1%2C16%5D%5C%5Cf%28A%29%20%5Ccap%20f%28B%29%20%3D%20%5B1%2C4%5D%20%20%3D%20f%28A%20%5Ccap%20B%29%5C%5C)
Therefore it is true for intersection. Now for union, we have that
![A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]](https://tex.z-dn.net/?f=A%20%5Ccup%20B%20%3D%20%5B0%2C4%5D%5C%5Cf%28A%5Ccup%20B%20%29%20%3D%20%5B0%2C16%5D%5C%5Cf%28A%29%20%3D%20%5B0%2C4%5D%5C%5Cf%28B%29%3D%20%5B1%2C16%5D%5C%5Cf%28A%29%20%5Ccup%20f%28B%29%20%3D%20%5B0%2C16%5D)
Therefore, for this case, it would be true that 
.
(b)
1 is not a set.
(c)
To begin with
Therefore
Now, given an element of 
it will belong to both sets, therefore it also belongs to 
, and you would have that
, therefore 
.
(d)
To begin with 
, therefore
