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Vesna [10]
3 years ago
14

Solve for x: (2/5)(x-4)=2x​

Mathematics
1 answer:
NARA [144]3 years ago
8 0

Simplify brackets

2/5(x - 4) = 2x

Simplify 2/5(x - 4) to 2(x - 4)/5

2(x - 4)/5 = 2x

Multiply both sides by 5

2(x - 4) = 10x

Divide both sides by 2

x - 4 = 5x

Subtract x from both sides

-4 = 5x - x

Simplify 5x - x to 4x

-4 = 4x

Divide both sides by 4

-1 = x

Switch sides

<u>x = -1</u>

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6 0
3 years ago
How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?
irga5000 [103]
The equivalence n \equiv 5 \pmod{12}

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5


for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)


we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009


The last k which produces n<5000 is 416


For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.


Answer: 417

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4 years ago
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