Answer:
#2. 3, #3. 9, #4. 6.2,$ 5. 9,# 6. 16, $7. 4, and #8. 3
Step-by-step explanation:
I hope this helps. Good Luck!
Answer:
Continuously
Step-by-step explanation:
Compounded continuously:
A = Pe^(rt)
A = 11,000 e^(0.0625 × 10)
A = 20,550.71
Compounded semiannually (twice per year):
A = P(1 + r)^t
A = 11,000 (1 + 0.063/2)^(2×10)
A = 11,000 (1 + 0.0315)^20
A = 20,453.96
Answer:
it is not a function because the input 5 (x) has more than one output (y)
domain (5,-1,8)
range (2,6,-2,3)
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
So I'm going to use vieta's formula.
Let u and v the zeros of the given quadratic in ax^2+bx+c form.
By vieta's formula:
1) u+v=-b/a
2) uv=c/a
We are also given not by the formula but by this problem:
3) u+v=uv
If we plug 1) and 2) into 3) we get:
-b/a=c/a
Multiply both sides by a:
-b=c
Here we have:
a=3
b=-(3k-2)
c=-(k-6)
So we are solving
-b=c for k:
3k-2=-(k-6)
Distribute:
3k-2=-k+6
Add k on both sides:
4k-2=6
Add 2 on both side:
4k=8
Divide both sides by 4:
k=2
Let's check:
:


I'm going to solve
for x using the quadratic formula:







Let's see if uv=u+v holds.

Keep in mind you are multiplying conjugates:



Let's see what u+v is now:


We have confirmed uv=u+v for k=2.