For f(x)=1/x^2-3
Find
A) f(3)
B) f(2-h)
If f(x)=1/x^2-3, then f(3) = 1 / 3^2 - 3. The exponentiation here must be carried out first: f(3) = 1/9 - 3. Then f(3) = 1/9 - 27/9 = -26/9
If f(x)=1/x^2-3, then f(2-h) = 1 / [2-h]^2 - 3. This result may be left as is or expanded. In expanded form, we have:
1
f(2-h) = ------------------ - 3
4-4h +h^2
Answer: $23
Step-by-step explanation:
1) 25-8=17
2) 17-4= 13
3) 13+10= 23
Matthew has $23 now
Step-by-step explanation:
For the quadratic equation to have 1 repeated real solution, the discriminant b² - 4ac must be zero.
=> (-z)² - 4(z - 5)(5) = 0
=> z² - 20(z - 5) = 0
=> z² - 20z + 100 = 0
=> (z - 10)² = 0
Therefore z = 10.
