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Pani-rosa [81]
3 years ago
11

Help with this math thingy pls! <3

Mathematics
1 answer:
zvonat [6]3 years ago
8 0
45.6 hope this helps
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Plz help I will give brainliest!
elena55 [62]

Answer:

b, 1 5/8

Step-by-step explanation:

the area is 4 15/32 square meters, and you find the area by doing length x width. the length is provided, 2 3/4. so all you have to do is divide 4 15/32 by 2 3/4 and you get 1 5/8

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Answer:

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Step-by-step explanation:

(10/y + 13) -3

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(10/5 + 13) -3

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6 0
2 years ago
Lucia draws a square and plots the center of the square. (image)
ss7ja [257]

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2 years ago
Could someone help me please 2x+5y=35
TEA [102]

• look below those are the options for y or x bc you didnt say which one we solving for.

3 0
2 years ago
Which summation formula represents the series below? 1 + 2 + 6 + 24
krek1111 [17]

Question:

Which summation formula represents the series below? 1 + 2 + 6 + 24

(a) \sum_{n=2}^{5}(n-1) !

(b) \sum_{n=0}^{3} n !

(c) \sum_{n=1}^{4}(n+1) !

(d) \sum_{n=2}^{5} n !

Answer:

Option a: \sum_{n=2}^{5}(n-1) ! is the correct answer.

Explanation:

Option a: \sum_{n=2}^{5}(n-1) !

By substituting the values of n and expanding the summation, we have,

(2-1) !+(3-1) !+(4-1) !+(5-1) !

Subtracting, we have,

1 !+2!+3 !+4 !

Expanding the factorial,

1+(2*1)+(3*2*1)+(4*3*2*1)

Simplifying, we get,

1+2+6+24

Thus, the summation \sum_{n=2}^{5}(n-1) ! represents the series 1+2+6+24

Hence, Option a is the correct answer.

Option b: \sum_{n=0}^{3} n !

By substituting the values of n and expanding the summation, we have,

0!+1!+2!+3!

Expanding the factorial,

0+1+(2*1)+(3*2*1)

Simplifying, we get,

0+1+2+6

Thus, the summation \sum_{n=0}^{3} n ! does not represents the series 1+2+6+24

Hence, Option b is not the correct answer.

Option c: \sum_{n=1}^{4}(n+1) !

By substituting the values of n and expanding the summation, we have,

(1+1) !+(2+1) !+(3+1) !+(4+1) !

Adding, we have,

2!+3!+4!+5!

Expanding the factorial,

(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)

Simplifying, we get,

2+6+24+120

Thus, the summation \sum_{n=1}^{4}(n+1) ! does not represents the series 1+2+6+24

Hence, Option c is not the correct answer.

Option d: \sum_{n=2}^{5} n !

By substituting the values of n and expanding the summation, we have,

2!+3!+4!+5!

Expanding the factorial,

(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)

Simplifying, we get,

2+6+24+120

Thus, the summation \sum_{n=2}^{5} n ! does not represents the series 1+2+6+24

Hence, Option d is not the correct answer.

Hence, the correct answer is Option a: \sum_{n=2}^{5}(n-1) !

6 0
3 years ago
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