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sweet-ann [11.9K]
3 years ago
7

Need help ASAP please

Mathematics
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

A- no

B- no

C-yes

D- yes

Step-by-step explanation:Anything that is inside the red is a solution. You can also plug in the points into the equation and see if its right. Thats what I did for a.

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Hello!
jarptica [38.1K]

Answer: The Answer is in a fraction form

Step-by-step explanation:

9/73

6 0
3 years ago
Read 2 more answers
Determine the first four terms of the sequence in which the nth term is
Licemer1 [7]

<u>Answer:</u>

The correct answer option is: \frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,\frac{1}{6}.

<u>Step-by-step explanation:</u>

We know that the nth term a_n for an arithmetic sequence is given by:

a_n=\frac{(n+1)!}{(n+2)!}

where n is the number of the position of the term.

We are supposed to find the first four terms of the sequence so we will substitute the values of n from 1 to 4 in the given formula to get:

1st term:

a_1=\frac{(1+1)!}{(1+2)!}=\frac{1}{3}

2nd term:

a_2=\frac{(2+1)!}{(2+2)!}=\frac{1}{4}

3rd term:

a_3=\frac{(3+1)!}{(3+2)!}=\frac{1}{5}

4th term:

a_4=\frac{(4+1)!}{(4+2)!}=\frac{1}{6}

8 0
3 years ago
Quadrilateral ABCD has the following
Daniel [21]

Answer:

Yes

Step-by-step explanation:

Because it’s a trapezoid and those can be trapezoid because they have one pair of parallel sides and a parallelogram has two pairs of parallel sides so a parallelogram is also a trapezoid.

4 0
3 years ago
Prove the identity.
EastWind [94]

Answer:

Hi... I should do like this

6 0
3 years ago
What are the three consecutive odd integers that equal 111? After that, answer this:Are there 3 consecutive odd integers that eq
ololo11 [35]
Odd integers are 1 more than even ones
even intergers are represetned as 2n where n is a whole number
odd=2n+1
consecutive odd integers are 2 apart from each other so therefor the numbers are 2n+1,2n+1+2, 2n+1+2+2 or 2n+1, 2n+3, 2n+5
so they add up to 111
2n+1+2n+3+2n+5=111
add like terms
6n+9=111
subtract 9 from both sides
6n=102
divide both sides by 6
n=17
numbers are
2n+1,2n+3,2n+5
subsitute
2(17)+1=34+1=35
therefor since the other sare 2 more
35,37,39 are the numbers


are there 3 that equal 1111? solve and see if it is true
2n+1+2n+3+2n+5=1111
add like terms
6n+9=1111
subtract 9 from both sides
6n=1102
divide both sides by 6
n=183.666666666666
n is not whole so the answer is NO

answer are 35,37,39 and NO
6 0
3 years ago
Read 2 more answers
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