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3 years ago

15

EMILY COLLECTED $950 SELLING GIRL SCOUT COOKIES ALL DAY SATURDAY. EMILY'S TROOP COLLECTED 10 TIMES AS MUCH AS SHE DID. HOW MUCH

MONEY DID EMILY'S TROOP RAISE?

2 answers:

Sauron [17]3 years ago

8 0

They made $9,500 MAN DATS A LOT OF MONEY $$$$$$$$

Katyanochek1 [597]3 years ago

7 0

9,500 just add the 0 always add 0 if you multiply by 10. 00 for 100 and 000 for 1000

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Evaluate the function when x=−2, 0, and 5. g(x)=3x−2

kipiarov [429]

I hope this helps....

x=-2
g(-2)= 3(-2)-2
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$\textit{Hello There!}$

$\text{\underline{ANSWER:}}$

$=3\sqrt{5} .$

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3 years ago

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Factor x3-2x2-9x+18 completely if(x+3) is a factor

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Factor x^3 -2x^2-9x+18

= x^2(x - 2) - 9(x - 2)

= (x^2 - 9)(x -2)

= (x + 3 )(x - 3)(x - 2)

So (x+3) is a factor

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4 years ago

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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

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4 years ago

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vova2212 [387]

Answer:

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Step-by-step explanation:

Line up the decimals and then add them up like usual, and then place the decimal where it should go. THen you get your answer.

Hope this helped!

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