Ann has seven balloons. Molly has a set less of balloons. How many balloons does molly have?

Mathematics

1 answer:

harina [27]3 years ago

4 0

0 balloons because if a set of balloons is 7 and Molly has a set less then it is 7-7=0

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Susie lost $15 on Monday. Her mom was upset, but gave her another $20. She lost another $13 today. Write an integer to represent

maria [59]

Answer:

Step-by-step explanation:

If you add 13 +15 you get 28$ if we were to write that as an integer out of 20 it would be 28/20 but as an improper fraction, it would need to be simplified into 1 and 8/20 ( 1 8/20)

If you are looking how much she lost out of the 20 she lost today it would be 13/20

Hope this helps

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4 0

3 years ago

Three consecutive integers have a sum of 42. find the integers.

ioda

The answer is 13,14,15.

6 0

3 years ago

What is the answer to this equation: 40+40-20x0+3+125=

MrRa [10]

Answer:

208

Step-by-step explanation:

20 × 0 = 0

40 + 40 = 80

125 + 3 = 128

3 0

3 years ago

Read 2 more answers

this year a farmer planted 300 acres of corn. this was 1.5 times as many acres as he planted last year. how many acres did he pl

diamong [38]

X • 1.5 = 300

So...

300/ 1.5 = 200

He planted 200 acres last year

6 0

3 years ago

Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1

nikklg [1K]

Given

$\Sigma_{n=1}^\infty15(-4)^n$

The first 10 partial sums are as follows:

$S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\ \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\ \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\ \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\ \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}$

$S_6=\Sigma_{n=1}^{6}15(-4)^n=\Sigma_{n=1}^{5}15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}$

The rest of the partial sums can be obtained in similar way.

7 0

3 years ago