Answer:
Roughly 2 years
Step-by-step explanation:
3500(1.05)^t=3850
(1.05)^t=1.1
t=1.953
So roughly 2 years
This means that you have to move the decimal point 4 decimal places to the left, as denoted by the negative sign. Thus, this would add 3 zeros before 5. The answer would be 0.00059.
I hope the explanation was clear to you. Have a good day.
Answer:
Fatima should shade 3 groups of 8 small squares or 24 small squares.
Step-by-step explanation:
Consider the provided information.
Fatima is shading this model to show 0.08 × 3. Shade the correct amount of boxes that will show the product.
Here 3 represents the number of groups and 8 represents the small square.
Multiplication is repeated addition.
We can rewrite the above multiplication as: 0.08 + 0.08 + 0.08
Therefore 0.08 × 3 = 0.24
Here 0.24 represents 24 small squares.
The required model is shown below:
Hence, Fatima should shade 3 groups of 8 small squares or 24 small squares.
Standard for is x × 10 ^I
where,
I = indice
10 > x < 0
you cannot write this is standard form with putting in what they stand for. However you can simplify it -
4(8m-7n) +6(3n-4m)
32m - 28n + 18n - 24m
8m - 10n
but hey I am from England do what we call standard form might be different
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
