All categories

3 years ago

Write the following comparison as a ratio reduced to lowest terms. 88 quarters to 30 dollars

1 answer:

6 0

You need to find common factors. The factors of 88 are 1 and 88, 2 and 44, 4 and 22, and 8 and 11. The factors of thirty are 1 and 30, 2 and 15, and 5 and 6. The greatest common factor is 2 so the ratio is 44:15.

You might be interested in

8_murik_8 [283]

It was a 15 percent tip

8 0

3 years ago

Read 2 more answers

Is 1/2x-4 a function?

lakkis [162]

Answer:

Yes it is

Step-by-step explanation:

Each x value is represented by exactly one y value.

7 0

3 years ago

Geometry help will give brainliest

Natali [406]

6x + 4x + 10 = 180
10x = 170
x = 17

mCA = 4(17) + 10 = 68 + 10 = 78

Answer:
mCA = 78°

8 0

3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago

What goes after 2, 9, 11, 18, 20, 27

Papessa [141]

Answer:

29 goes after 27

Step-by-step explanation:

the patern is add 2, then add 7, repeat

7 0

3 years ago

Read 2 more answers