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Verizon [17]
3 years ago
7

What is the best first step for solving the given system using substitution while avoiding fractions

Mathematics
1 answer:
Eduardwww [97]3 years ago
6 0

Hi, hope you’re having a good day.

You should probably add an attachment.

Please do! Thanks, Have a great day! ❤️❤️

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rectangle A has a length of 24 meters and a width of 20 meters. rectangle B has the same area as rectangle A. the width of recta
blagie [28]
To solve this, first we'll find the area of the rectangle A,

Area=length × width
?=24m×20m
480m=24m×20m
480m squared=area of the rectangle A

now we'll find the width of rectangle B,
"the width of rectangle B is 12 meters less than the width of rectangle A",
20m-12m= 8m
8m=width of rectangle B

finally we'll find the length of rectangle B,
area of the rectangle B= 480msquared
width= 8m
length=? (to find this divide the area by the width)
480÷8=60m
length of the rectangle B=60m



7 0
4 years ago
Why is it so hard for people to learn math?
ddd [48]

Answer:

I don't know, I kind of wonder the same thing sometimes. But everyone understands things at a different rate, so that's why I'm here helping people by answering their questions and doing my best to explain how I got there.

thank you for the free points though :)

6 0
3 years ago
Read 2 more answers
Jessie says to round 763,400 to the nearest ten thousand, he will round to 770,000. is he right? Explain
bogdanovich [222]
763,400 rounded to the nearest ten thousand = 760,000....Jessie is wrong
4 0
3 years ago
Read 2 more answers
Please HELP!!!
anygoal [31]
I think the answer is (3,3) I’m sorry if it’s wrong
5 0
3 years ago
Prove that sequence an = (3^n)/(2n)! is monotone
krek1111 [17]
When n=1, you have a_1=\dfrac3{2!}=\dfrac32.

When n=2, you have a_2=\dfrac{3^2}{4!}=\dfrac38.

Clearly, a_1>a_2.

Assume a_k. Now when n=k+1, you have

a_{k+1}=\dfrac{3^{k+1}}{(2k+2)!}=\dfrac3{(2k+2)(2k+1)}\times\dfrac{3^k}{(2k)!}=\dfrac3{(2k+2)(2k+1)}a_k

Therefore by induction the sequence is monotone (decreasing).
5 0
3 years ago
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