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3 years ago

Find the least common multiple of 20, 4, and 5

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

LCM = 20

Step-by-step explanation:

You need to make everything into 20. Luckily, 4X5 is 20 and vise versa.

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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

I'm making a table.

Ivenika [448]

Well I think I can help you out ...
if you assume input x=2
the output will be as following :
y= -1/3(2)+2
=-2/3+2
=-2/3+6/3
=(-2+6)/3
=4/3
=1.333

3 0

4 years ago

Solve the following equation using the quadratic formula.

nataly862011 [7]

Answer:

x=3.5,x=-6

I hope this help

7 0

3 years ago

Please help it is due tomorrow.

svlad2 [7]

R = radius of basketball
r = radius of tennis ball
R = 3.6r cube both sides
the ratio of the radii is 1: 3.6
cube both sides to find the ratio of the volumes
1 : 46.656 ( this many times greater)

6 0

3 years ago

You need gas for your car, and just as you are approaching where you normally get gas, the radio station announces that the Chea

schepotkina [342]

Answer: Yes

Step-by-step explanation:

Given that:

Distance from present location to CheapGas = 4 miles

Price of gas at CheapGas is $0.37 cheaper

Hence, the decision to make a U-turn back to CheapGas or continue on intended destination will rely on two factors ;

1) The number of gallons of gas to purchase

2) The cost of gas required to return to cheap gas.

If a substantial amount of gas is to be purchased, may be 50, 100 or more gallons, the amount saved by making a turn back to cheapGas and purchase gas at $0.37 cheaper per gallon.

8 0

3 years ago