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svetoff [14.1K]
3 years ago
11

2x-1/6 = 2(x-3)/3 + 1?

Mathematics
2 answers:
IceJOKER [234]3 years ago
7 0
2x-\frac{1}{6}=\frac{2(x-3)}{3}+1\ \ \ \ |multiply\ both\ sides\ by\ 6\\\\6(2x)-6(\frac{1}{6})=6\left(\frac{2x-6}{3}\right)+6(1)\\\\12x-1=2(2x-6)+6\\\\12x-1=2(2x)+2(-6)+6\\\\12x-1=4x-12+6\\\\12x-1=4x-6\ \ \ \ |add\ 1\ to\ both\ sides\\\\12x=4x-5\ \ \ \ \ |subtract\ 4x\ from\ both\ sides\\\\8x=-5\ \ \ \ |divide\ both\ sides\ by\ 8\\\\\boxed{\boxed{x=-\frac{5}{8}}}
Crazy boy [7]3 years ago
5 0
<u>2x - 1</u> = <u>2(x-3)</u>
   6        3 + 1
<u>2x - 1</u> = <u>2x - 6</u>
   6            4
4(2x - 1) = 6(2x-6)
8x - 8 = 12x - 36
<u>-8x        -8x        </u>
     -8 = 4x - 36
<u>  +36         +36</u>
    28 = 4x
    <u>28</u> = <u>4x</u>
     4      4
     7 = x
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Ksenya-84 [330]
To obtain the square root of 16x^36, the coefficient portion (16) will not present any problems since 16 is a perfect square. However, for a variable with an exponent, the exponent is to be multiplied by 1/2 since the square root symbol is equal to raising the term inside to the power of 1/2. This is shown below:

sqrt (16 x^36) = 4 * x^36(1/2) = 4 * x^18

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3 years ago
List all x intercepts for y= -cos(1/2x + 1/5 pi) on the interval [-2/5pi,4pi]
Dahasolnce [82]
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