The smallest positive integer n for which n^2 is divisible by 18 and n^3 is divisible by 640 is 120
The modulo (or "modulus" or "mod") is the remainder after dividing one number by another.
The smallest integer is 120, since:
mod 18 = 0, (reminder after dividing
by 18 is 0) and
mod 640 = 0 (reminder after dividing
by 18 is 0)
Hence, The smallest positive integer n for which n^2 is divisible by 18 and n^3 is divisible by 640 is 120.
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Answer:
D.
Step-by-step explanation:
adding 8 cancels it out on side with x, the left side
In point slope form:
y+3=1/2(x-2)
In slope intercept
y= 1/2x-4
Answer:
x=−2
y=7
Step-by-step explanation:
5x+2y=4
x−3y=−23
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
5x+2y=4,x−3y=−23
To make 5x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 5.
5x+2y=4,5x+5(−3)y=5(−23)
Simplify.
5x+2y=4,5x−15y=−115
Subtract 5x−15y=−115 from 5x+2y=4 by subtracting like terms on each side of the equal sign.
5x−5x+2y+15y=4+115
Add 5x to −5x. Terms 5x and −5x cancel out, leaving an equation with only one variable that can be solved.
2y+15y=4+115
Add 2y to 15y.
17y=4+115
Add 4 to 115.
17y=119
Divide both sides by 17.
y=7
Substitute 7 for y in x−3y=−23. Because the resulting equation contains only one variable, you can solve for x directly.
x−3×7=−23
Multiply −3 times 7.
x−21=−23
Add 21 to both sides of the equation.
x=−2
The system is now solved.
x=−2,y=7
Graph if needed: