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2 years ago

Why is the hypotenuse always the longest side of a triangle?

Mathematics

1 answer:

yawa3891 [41]2 years ago

4 0

Because you have to connect the other two sides , so a diagonal line will be longer than a vertical line

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An army base had provisions for 300 soldiers for

alukav5142 [94]

Answer:

Step-by-step explanation:

To find out how much we will divide 90 by 300

90 ÷ 300 = 0.3

Then after subtracting 50 soldiers = 300 - 50 = 250 soldiers

Then 250 x 0.3 = 75 days

5 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

The cost of oranges varies directly as the number of oranges sold. If 8 oranges cost $10.40, how many oranges can be bought for

alexdok [17]

Answer:

Step-by-step explanation:

if you can get 8 oranges for 10.40 then divide 10.40 by 8. you can get each orange for 1.3. you need to know how many oranges you can get for 33.8 so divide that by 1.3 and you get 26

6 0

3 years ago

Read 2 more answers

Is y=3x-5 and 6x=2y+10 a solution

iVinArrow [24]

Y=3x-5
y-3x=-5
Times 2 from both side
y(2)-3x(2)=-5(2)
2y-6x=-10
Times negative from both side
(-)2y-6x(-)=(-)-10
-2y+6x=10
or
6x-2y=10

6x=2y+10
6x-2y=10
Furthermore, we see that both equation have the same 6x-2y=10 which means that both of them have infinity solutions, not a solution. Hope it help!

5 0

3 years ago

The area of a rectangle is (27x5 + 9x4 - 18x3) square feet. The width of the rectangle is 9x2 feet. What is the length of the re

Yuliya22 [10]

If the width is 9x², then the length is 27x^5+9x^4-18x^3/9x², which equals 3x³+x²+2x. ☺☺☺☺

8 0

3 years ago

Read 2 more answers