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2 years ago

Three consecutive integers have a sum of 75. Find the integers

Mathematics

2 answers:

natima [27]2 years ago

8 0

The answers is 24, 25, and 26.

qaws [65]2 years ago

6 0

Answer: 24, 25, 26

Step-by-step explanation:

Three consecutive integers can be written as :

(x),(x+1),(x+2)

Adding them: (x)+(x+1)+(x+2)=75

3x+3=75

3x=75-3

=72

3x=72

x=24

so, x+1=25, x+2=26

the numbers are

24, 25, 26,

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A = 15 ; b =____________ ; c = 17

Lorico [155]

Answer:

Step-by-step explanation:

b=16

Following the pattern, we can assume that each term adds by 1. So 15+1 = 16.

3 0

2 years ago

In a right triangle the measure of one acute angle is two times the sum of the measure of the other acute angle and 6. find the

Llana [10]

180 - 90 = 90.

x + 2x = 90
3x = 90
x = 30

One acute angle is 30, the other is 60.

8 0

2 years ago

Which rectangles have the same perimeter?

myrzilka [38]

A and B have the same perimeter

8 0

2 years ago

Read 2 more answers

Find the mean for the amounts: $17,485; $14,978; $13,592; $14,500; $18,540; $14,978. Round to the nearest dollar.

Phoenix [80]

Answer:$15,679

Step-by-step explanation:

Mean is calculated as total sum of the amount divided by number of the amount.

From the given question, we add the amount together and divide by six:

Mean = $17,485 + $14,978+ $13,592 +$14,500 + $18,540 +$14,978/6

=$94,073/6

=$15,678.8

To the nearest dollar will be $15,679 since 0.8 is greater than 0.5, we round up.

I hope this helps.

8 0

3 years ago

Describe the rule for the sequence 2, 1, 2/3, 1/2, 2/5, 1/3, 1/7,...

Triss [41]

Multiply 2 by 1/2 to get 1.

Multiply 1 by 2/3 to get 2/3.

Multiply 2/3 by 3/4 to get 6/12 = 1/2.

Multiply 1/2 by 4/5 to get 4/10 = 2/5.

Multiply 2/5 by 5/6 to get 10/30 = 1/3.

Multiply 1/3 by 6/7 to get 6/21 = 2/7. (I suspect there's a typo in the question.)

And so on, so that the nth term in the sequence is multiplied by n/(n + 1) to get the (n + 1)th term.

Recursively, the sequence is given by

$\begin{cases}a_1=2\\a_n=\dfrac{n-1}na_{n-1}&\text{for }n>1\end{cases}$

We can solve this exactly by iterating:

$a_n=\dfrac{n-1}na_{n-1}=\dfrac{n-1}n\dfrac{n-2}na_{n-1}=\dfrac{n-1}n\dfrac{n-2}{n-1}\dfrac{n-3}{n-2}a_{n-3}=\cdots$

and so on down to

$a_n=\dfrac{(n-1)\cdot(n-2)\cdot(n-3)\cdot\cdots\cdot3\cdot2\cdot1}{n\cdot(n-1)\cdot(n-2)\cdot\cdots\cdot4\cdot3\cdot2}a_1$

$a_n=\dfrac{(n-1)!}{n!}a_1$

and with lots of cancellation, we end up with

$a_n=\dfrac{a_1}n=\boxed{\dfrac2n}$

6 0

2 years ago