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saul85 [17]
2 years ago
13

What is math forlike what do we use it for whats the point of math no one uses it

Mathematics
1 answer:
wolverine [178]2 years ago
6 0
Because,mathematics provides an effective way of building mental discipline and encourages logical reasoning and mental rigor. In addition, mathematical knowledge plays a crucial role in understanding the contents of other school subjects such as science, social studies, and even music and art.
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Solve the equation <br> 3|x – 2|+2 = 17
grigory [225]

Answer:

-3, 7 = x

Step-by-step explanation:

3|x – 2| + 2 = 17

- 2 - 2

________________

3|x - 2| = 15

_______ ___

3 3

|x - 2| = 5

x - 2 =  -5 \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  x - 2 = 5

+ 2 + 2 + 2 + 2

_______________ ____________

x = -3 x = 7

I am joyous to assist you anytime.

5 0
3 years ago
PLEASE HELP!!!!! 15 PTS
zvonat [6]
Quoteint of powers
(x^m)/(x^n)=x^(m-n)

we know that 8=x^3

so
(2^5)/8=2^2 can be rewritten as
(2^5)/(2^3)=2^2
and 5-3=2 so it's true

answer is
third one
by simplifieng 8 to 2^3 to make both powers base two, and subtraction the exponents
7 0
3 years ago
Read 2 more answers
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
If f(x)=-9x-3, find f(-5)
Andrews [41]

Answer:

42

Step-by-step explanation:

f(-5) = -9 × (-5) - 3 = 45 - 3 = 42

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below =)

5 0
3 years ago
In how many ways can a committee of four men and four woman be formed from a group of seven men and eleven women?
-Dominant- [34]
Choosing 4 men from seven can be done in 7C4 = 35 ways
(There are 7 to choose from then 6 and 5 and 4 = 7 x 6 x 5 x 4
but we don’t want them in order so we need to divide by 4! = 4 x 3 x 2 x 1)
Similarly choosing 4 women from 11 can be done in 11C4 = 330 ways.

Total number of ways to choose 4 men and 4 women = 35 x 330 = 11550
3 0
1 year ago
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