1.
If $x <> (sqrt xy)/(|x-y|), find
4 <> (9 <> 1) to the nearest tenth.
<>=diamond
This is a function, where the "diamond" represents a function.
2.
If f(x) = 3x^2 - 5, what is the value of f(f(-1))?
If I read this right,
x<>y = √(xy)/|x-y|
so,
9<>1 = √(9*1)/|9-1| = 3/8
4<>(3/8) = √(4*3/8)/|4-3/8| = √(3/2)/(31/8) = 4/31 √6
f(-1) = 3*1-5 = -2
f(-2) = 3*4-5 = 7
so, f(f(-1)) = 7
4-3/8 = 29/8, not 31/8
1.
To find the value of 4 <> (9 <> 1) to the nearest tenth, we need to evaluate the expression inside the diamond function first. Let's focus on (9 <> 1) first.
In this case, the diamond function represents a function where the first value in the parentheses is subtracted from the second value. So, (9 <> 1) means subtracting 9 from 1:
1 - 9 = -8
Now, we can substitute this result back into the original expression:
4 <> -8
Similarly, the diamond function subtracts the second value from the first value. So, 4 - (-8) can be simplified to:
4 + 8 = 12
Therefore, the value of 4 <> (9 <> 1) to the nearest tenth is 12.
2.
To find the value of f(f(-1)), we need to first evaluate f(-1) and then substitute the result into f(x) function.
Given f(x) = 3x^2 - 5, we substitute -1 into f(x):
f(-1) = 3(-1)^2 - 5
= 3(1) - 5
= 3 - 5
= -2
Now, substitute -2 back into f(x) function:
f(f(-1)) = f(-2)
= 3(-2)^2 - 5
= 3(4) - 5
= 12 - 5
= 7
Therefore, the value of f(f(-1)) is 7.