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kolezko [41]
2 years ago
10

The smallest of three consecutive integers is 24 less than the sum of the two larger integers. Find the three integers.

Mathematics
1 answer:
harkovskaia [24]2 years ago
3 0

Answer:

21, 22, 23

Step-by-step explanation:

Construct equations representing each piece of information.

Three consecutive integers are three numbers in a row, for example, 1, 2 and 3.

Let's let

Integer 1 = X

Integer 2 = Y

Integer 3 = Z

X = (Y + Z) - 24

We also know that

Y = X + 1

and Z = X + 2

We can now sub this information into the first equation. We do this so that we can have only one variable in the equation.

X = (X+1) + (X+2) - 24

now solve for x,

X = 2X + 3 - 24

X = 2X - 21

-X = - 21

X = 21

So, the integers are 21, 22, 23

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shepuryov [24]

Answer:

48 cups

Step-by-step explanation:

16 times 3

7 0
3 years ago
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Frank says that for any whole number n, the value of 6n-1 is always prime. Is Frank correct? Explain your answer.​
Lady bird [3.3K]

Answer:

Frank is wrong

Step-by-step explanation:

6(6) - 1 = 35

35 is a multiply by 5 so it is not a prime number.

8 0
3 years ago
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lisov135 [29]

Answer:

A i think

Step-by-step explanation:

8 0
2 years ago
Given that the series kcoskt kº +2 k=1 converges, suppose that the 3rd partial sum of the series is used to estimate the sum of
3241004551 [841]

Answer:

c

Step-by-step explanation:

Given that:

\sum \limits ^{\infty}_{k=1} \dfrac{kcos (k\pi)}{k^3+2}

since cos (kπ) = -1^k

Then, the  series can be expressed as:

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^kk)}{k^3+2}

In the sum of an alternating series, the best bound on the remainder for the approximation is related to its (n+1)^{th term.

∴

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(3+1)}(3+1))}{(3+1)^3+2}

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(4)}(4))}{(4)^3+2}

= \dfrac{4}{64+2}

=\dfrac{2}{33}

5 0
2 years ago
How many lb of brand X sugar which
Dafna1 [17]

Answer:  <u>4 pounds</u> of brand X sugar

====================================================

Reason:

n = number of pounds of brand X sugar

5n = cost of buying those n pounds, at $5 per pound

Brand Y costs $2 per pound, and you buy 8 lbs of it, so that's another 2*8 = 16 dollars.

5n+16 = total cost of brand X and brand Y combined

n+8 = total amount of sugar bought, in pounds

3(n+8) = total cost because we buy n+8 pounds at $3 per pound

The 5n+16 and 3(n+8) represent the same total cost.

Set them equal to each other. Solve for n.

5n+16 = 3(n+8)

5n+16 = 3n+24

5n-3n = 24-16

2n = 8

n = 8/2

n = 4 pounds of brand X sugar are needed

-------------

Check:

n = 4

5n = 5*4 = 20 dollars spent on brand X alone

16 dollars spent on brand Y mentioned earlier

20+16 = 36 dollars spent total

n+8 = 4+8 = 12 pounds of both types of sugar brands combined

3*12 = 36 dollars spent on both types of sugar brands

The answer is confirmed.

--------------

Another way to verify:

5n+16 = 3(n+8)

5*4+16 = 3(4+8)

20+16 = 3(12)

36 = 36

6 0
2 years ago
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