This question is Incomplete
Complete Question
Recall the hurricane relief fund context from the previous investigation. Bob set up a hurricane relief donation fund and started it off by donating $ 35 . Instead of tripling in value each day, suppose it doubled in value each day
Complete the following table showing the account values at the end of each day.
Number of days since the initial investment, n Amount (in dollars) in the relief account, A
0 1 2 3 4
Answer:
Number of days(n) = Amount (in dollars) in the relief account(A)
0 = $35
1 = $70
2 = $140
3 = $280
4 = $560
Step-by-step explanation:
The formula for doubling time
P(t) = Po(2)^t/k
Where P = Amount after time t
Po = Initial Amount
t = number of days
k = Frequency at which it doubles
For 0 days
t = 0, k = 1, P(o) = $35
P(t) = Po(2)^t/k
P(0) = $35(2)^0/1
= $35(2)⁰
= $35 × 1
= $35
For 1 day
t = 1, k = 1 , P(0) = $35
P(t) = Po(2)^t/k
P(0) = $35(2)^1/1
= $35(2)¹
= $35 × 2
= $70
For 2 days
P(t) = Po(2)^t/k
t = 2, k = 1, P(0) = $35
P(0) = $35(2)^2/1
= $35(2)²
= $35 × 4
= $140
For 3 days
P(t) = Po(2)^t/k
t = 3, k = 1
P(0) = $35(2)^3/1
= $35(2)³
= $35 × 8
= $280
For 4 days
P(t) = Po(2)^t/k
t = 3, k = 1, P(0) = $35
P(0) = $35(2)^5/1
= $35(2)⁴
= $35 × 16
= $560
All points in the shaded area are answers to the inequality. Thus you need to find the point which is in the shaded area of the graph.
Only the point (3, -1) is within range. Thus answer D is correct.
Answer:
27 years
Step-by-step explanation:
Given that :
Adrian's age = 3
Father's age = 34
In how many years will Adrian be twice as young as his father?
Let the number of years = x
2(3 + x) = 34 + x
6 + 2x = 34 + x
2x - x = 34 - 6
x = 28
Answer: 0.79
...................................................
Answer:
, ∠C ≅∠Z
, ∠A≅∠X

Step-by-step explanation:
we know that
If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent
so
In this problem
The corresponding sides are
BA and YX
BC and YZ
AC and XZ
The corresponding angles are
∠A and ∠X
∠B and ∠Y
∠C and ∠Z
so

and
∠A≅∠X
∠B≅∠Y
∠C ≅∠Z
therefore
, ∠C ≅∠Z
, ∠A≅∠X
