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Makovka662 [10]
3 years ago
15

Catherine and Lance open a savings account at the same time. Catherine deposits $100 initially and adds $20 per week. Lance depo

sits $500 initially and adds $10 per week. Catherine wants to know when she will have the same amount in her savings account as Lance.
Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
5 0

Answer:

after 40 weeks

Step-by-step explanation:

docker41 [41]3 years ago
4 0

Answer:

After 40 weeks

Step-by-step explanation:

To get the answer to this question, we need to first know the difference in the amount of money deposited by the 2 individuals and in this case it us $400 since one deposited $500 and the other $100

The next thing to do is to divide the difference in amount(400) by the amount contributed by each person per week.

Catherine deposits $20 per week

400/20 = 20

Lance deposits $10 per week

400/10 = 40

The LCM between 40 and 20 is 40 weeks.

Therefore, both individuals need to deposit money for 40 weeks for both of their savings to be the same.

If Catherine deposits $20 for 40 weeks,she will have 40 × 20 = $800 + $100(initial deposit that she made) = $900

Lance deposits $10 for for 40 weeks and he will have 40 × 10 = $400 + $500(initial deposit made by him)= $900

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for
KIM [24]

Answer:

a) The recurrence formula is P_n = \frac{109}{100}P_{n-1}.

b) The general formula for the population of Tacoma is

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write P_0 = 200 000. For the population in 2001 we will use P_1, for the population in 2002 we will use P_2, and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that P_0 is equal to P_1 plus 9% of the population of 2000. Notice that this can be written as

P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0.

In 2002, we will have the population of 2001, P_1, plus the 9% of P_1. This is

P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1.

So, it is not difficult to notice that the general recurrence is

P_n = \frac{109}{100}P_{n-1}.

b) In the previous formula we only need to substitute the expression for P_{n-1}:

P_{n-1} = \frac{109}{100}P_{n-2}.

Then,

P_n = \left(\frac{109}{100}\right)^2P_{n-2}.

Repeating the procedure for P_{n-3} we get

P_n = \left(\frac{109}{100}\right)^3P_{n-3}.

But we can do the same operation n times, so

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) Recall the notation we have used:

P_{0} for 2000, P_{1} for 2001, P_{2} for 2002, and so on. Then, 2016 is P_{16}. So, in order to obtain the approximate population of Tacoma in 2016 is

P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062

d) In this case we want to know when P_n>400000, which is equivalent to

(1.09)^{n}P_0>400000.

Substituting the value of P_0, we get

(1.09)^{n}200000>400000.

Simplifying the expression:

(1.09)^{n}>2.

So, we need to find the value of n such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

n\ln(1.09)>\ln 2.

So, n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693.

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0
3 years ago
Read 2 more answers
Which answer is correct? It wouldn’t let me click mathematics btw
Bad White [126]
6)
KM bisects on <NKL so m<MKN = m<MKL = 3x + 5

2(m<MKL) + m<JKN = 180
2(3x+5) + 8x + 2 = 180
6x + 10 + 8x + 2 = 180
14x + 12 = 180
14x = 168
    x = 12

so 
 m<MKN = m<MKL = 3x + 5 = 3(12) + 5 = 36 + 5 = 41

answer
b. 41

7)

2 supplementary has sum = 180
x + x + 14 = 180
2x + 14 = 180
2x = 166
  x = 83
  x + 14 = 83 + 14 = 97

2 angles are 83 and 97

answer
b. 83, 97
7 0
4 years ago
HELP ME PLEASEEEEEEE
scoray [572]

Answer:

It would be in this order:

Types of Quads:

Rectangle

Square

Rhombus

Trapezoid

Type of Triangles:

Equaliteral

Isoceles

Scalene

Step-by-step explanation:

Hope it helps

6 0
3 years ago
IMPORTANT PLEASE I NEED NOW:Find the area of each face of the cube whose volume is 729 cm
vivado [14]

Answer:

81

Step-by-step explanation:

square root of 729= 9

9×9=81

8 0
3 years ago
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Find the perimeter of a rectangle that is 2 and 1/2 yards by 4 and 1/2 yards
Travka [436]
The answer is 14. First, get both numbers and add them twice (2.5+2.5+4.5+4.5) or (2.5 * 2 + 4.5 * 2). Then you should get 2 numbers which are 9 and 5. Then it should be 14, which is the answer.
7 0
3 years ago
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