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3 years ago

Helpppppppp meeeeeeeeeeeeeeeeeeee

Mathematics

1 answer:

Alex73 [517]3 years ago

6 0

Answer:

The length of the shorter piece is $5\textrm{ ft}$ .

The length of the longer piece is $11\textrm{ ft}$ .

Step-by-step explanation:

The five-step method is GUESS method.

Given:

Length of steel beam, $L=16\textrm{ ft}$

Length of longer piece is 1 foot longer than twice the shorter one.

Unknown:

Length of longer and shorter pieces of beam.

Let the length of shorter piece be $x$ feet.

Therefore, the length of longer beam is equal to $1+2x$

Evaluate:

Sum of lengths of longer and shorter pieces of beam is equal to the total length of beam. So,

$x+1+2x=16\\3x+1=16\\3x=16-1\\3x=15\\x=\frac{15}{3}=5$

Substitute and Solve:

Now, we have, $x=5$ .

Substitute 5 for $x$ for the lengths of shorter and longer pieces and solve for each. This gives,

Length of shorter piece = $x=5\textrm{ ft}$

Length of longer piece = $1+2x=1+2(5)=1+10=11\textrm{ ft}$

Therefore, the lengths of shorter and longer pieces are $5\textrm{ ft}$ and $11\textrm{ ft}$ respectively.

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Answer:

Answer:

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Step-by-step explanation:

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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

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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