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kirill [66]
3 years ago
11

How to solve these kind of questions.

Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0
To solve those you need to FOIL or rainbow the distributive property

hope this helped
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I need the answer asap .! help please !
shtirl [24]

Answer:

i tried the math and if im correct it should be 0.52N

Step-by-step explanation:

im sorry if its not correct bro

8 0
3 years ago
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
Hi can someone please help me?
jasenka [17]
Evaluate.
1934917632
this is evaluation
8 0
3 years ago
Read 2 more answers
Write three different sentences that could describe the relationship between the quantities $27, $81
Olenka [21]
1) 27+x=81
2)27+81=108
3)81-x=27
8 0
3 years ago
If EG = 59, and EF = 8x - 14, while FG = 4x + 1
lyudmila [28]

Answer:

x = 6

Step-by-step explanation:

EG = 59

EF = 8x - 14

FG = 4x + 1

to find the value of x

EF + FG = EG (same line)

8x - 14 + 4x + 1 = 59

12x -13 = 59

12x = 59 + 13

12x = 72

x = 72 ÷ 12

x = 6

You can even cross check for correct answer

EF + FG = EG

instead of x place the x's value we got

8x - 14 + 4x + 1 = 59

8 (6) - 14 + 4 (6) + 1 = 59

48 - 14 + 24 + 1 = 59

34 + 25 = 59

59 = 59

CROSS - CHECKED

4 0
3 years ago
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