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maw [93]
3 years ago
5

find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

11, 13

Step-by-step explanation:

Let n and n+2 be the consecutive odd integers

2(n+2) = 3n + 15

2n + 4 = 3n + 15

n = 11

n + 2 = 13

Svetach [21]3 years ago
4 0

Answer:

11, 13

Step-by-step explanation:

an odd number can be represented by 2n+1

since they are consecutive, the larger odd number will be 2n + 1 + 2

now,

2(2n+3) = 2n + 1 +15

solving this eqn, we get n = 5

so the two numbers are (2n+1) = 11 and 11+2= 13

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3 years ago
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Step-by-step explanation:

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2 years ago
How many solutions does 9(z+8)=-9z-72 have
zloy xaker [14]

Answer:

1 solution.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

9(z + 8) = -9z - 72

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Distribute 9:                              9x + 72 = -9z - 72
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Here we see that we will get only 1 solution for <em>z</em>.

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