Help with this math thingy pls! <3

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

6 0

2 years ago

Read 2 more answers

HELP PLEASE ASAPPPP!!

MakcuM [25]

Answer:

12

Step-by-step explanation:

(10/y + 13) -3

Let y=5

(10/5 + 13) -3

PEMDAS says parentheses first

(2 +13) -3

15 -3

12

6 0

2 years ago

Lucia draws a square and plots the center of the square. (image)

ss7ja [257]

Answer:

C

Step-by-step explanation:

Lucia's claim is correct since any rotation that is a multiple of 45° carries a square onto itself

8 0

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