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erik [133]
2 years ago
10

The sum of two consecutive integers is 65. what are the two integers? show steps please.

Mathematics
1 answer:
Gnoma [55]2 years ago
7 0

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the two integers, given that they are consecutive, and their sum is 65.

\triangle~\fbox{\bf{KEY:}}

  • Consecutive integers are right next to each other, like 12 and 13. or 65 and 66.

Let the first integer be x, and let the second integer be x+1.

Their sum is 65. Let's set up our equation:

\longmapsto\sf{x+x+1=65}

Combine like terms:

\longmapsto\sf{2x+1=65}

Subtract 1 from both sides of the equal sign:

\longmapsto\sf{2x=64}

Divide both sides by 2:

\longmapsto\sf{x=32}

To find the second integer, subtract the first integer from the sum of the two integers:

\longmapsto\pmb{65-32}

\longmapsto\pmb{33}

The integers are: 33 and 32.

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

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Find the length of QP given Q is the midpoint of XF, PQ =2X+1, XF=7X-4, PF=X
nadezda [96]

Answer:

Therefore the length of QP = 3.4 units

Step-by-step explanation:

Given:

PQ = 2x + 1

XF = 7x - 4

PF = x

Q is the mid poimt of XF

∴ XQ = QF

QF = PQ - PF  ..........( Q - F - P )

     = 2x + 1 - x

∴ QF      = x + 1

∴ XQ = QF = x + 1

TO Find:

QP = ?

Solution:

By Addition Property we have

XP = XQ + QF+PF ..........(X-Q-F-P)\\\\

XF + PF =XQ + QF+PF ..........(X-Q-F-P)\\

Substituting the given values in above equation we get

(7x - 4) + x = (x +1) + (x +1) + x

8x -4 = 3x +2

8x - 3x + 4 + 2

5x = 6

∴ x = \frac{6}{5}

Now we require

QP = (2x + 1)

∴ QP = 2\times \frac{6}{5} +1\\\\QP = \frac{12+5}{5} \\\\QP =\frac{17}{5} \\\\\therefore QP = 3.4\ unit

Therefore the length of QP = 3.4 units

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