1. As x approaches 1, f(x) approaches 3-1, that is 2.
If a = 2 and b = 3, f(1) = 2(1)^2 + 3(1) = 5
So there is a 'jump' in values of x at x = 1. So its not continuous at x=1.
2. For continuity at x = 1, ax^2 + bx must = 2 that is when a + b = 2.
3. for continuity at x = 2 , ax^2 + bx must be = 0 that is when a*2^2 + 2b = 0
- that is 4a + 2b = 0.
4. we have system of equations:-
a + b = 2
4a + 2b = 0
this gives a = -2 and b = 4 So f(x) is continuous when a = -2 and b = 4.
Yes. The two smaller squares have a sum of 169 which is the value of the larger square.
a^2 + b^2 = c^2
25 + 144 = 169
It can be done. Notice the figure below shows you how to arrange the squares to give the answer of a^2 + b^2 = c^2
Answer:
Step-by-step explanation:
one-hundred and ten-thousand and eleven
Answer:
a.) 0.83
Step-by-step explanation:
