All categories

3 years ago

Simplify all this AND SHOW YOUR WORK

Mathematics

2 answers:

Elena L [17]3 years ago

7 0

Answer:

4/10 = 2/5

18/24 = 6/8 = 3/4

8/20 = 4/10 = 2/5

3/8 + 3/8 = 6/8 = 3/4

8/9 - 2/9 = 6/9 = 2/3

Sliva [168]3 years ago

4 0

Answer:

Step-by-step explanation:

Answer:

4/10 = 2/5

18/24 = 6/8 = 3/4

8/20 = 4/10 = 2/5

3/8 + 3/8 = 6/8 = 3/4

8/9 - 2/9 = 6/9 = 2/3

You might be interested in

Priya tried to solve an equation step by step. \qquad\begin{aligned} \dfrac f{0.25}&=16\\\\ \\ \dfrac{f}{0.25} \cdot0.25&amp

Zarrin [17]

Answer:

f=4

Step-by-step explanation:

Priya Tried to solve the equation below step-by-step:

$\qquad\begin{aligned} \dfrac f{0.25}&=16\\ \dfrac{f}{0.25} \cdot0.25&=16\cdot0.25&\green{\text{Step } 1}\\ f&=4&\blue{\text{Step } 2}\\\end{aligned}$

The steps are correct and indeed f=4.

7 0

3 years ago

Read 2 more answers

What combination of transformations is shown below?

saw5 [17]

Answer:

Rotated then reflected.

Step-by-step explanation:

3 0

3 years ago

Please help. Will reward brainliest

Mashutka [201]

Answer:

C is the answer for this one

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