When positive integer N is divided by 167, the remainder is 35, and when positive integer K is divided by 167, the remainder is 17.
What is the remainder when 2*N+K is divided by 167?_____
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$$\sqrt{(12*32*54)}$$
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What is the decimal equivalent of $$(\frac{2}{5})^5$$ ?
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If 9^(2*x+5) = 27^(3*x-10), then x =
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If f(x) = x^4 - 3*(x^3)- 2*(x^2) + 5*x , then f(-1) =
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sqrt[(49*137)- (56*49)]=
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(9^k)*[27^(2*k)]=
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(73^2)+(74^2)=
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[3*(10^20)]*[3*(10^(-5))]=
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[2*x*(y^2)]*[7*(x^3)*(y^3)]=
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$$8·(10^{40})\over1·(10^{10})$$
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If (10^a)*(10^b)*(10^c) = 1,000,000, and a, b, and c are different positive integers, then 10^a + 10^b + 10^c =
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[3*(10^20)]*[8*(10^30)]=
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{[3^(-1)]-[2^(-1)]}^(-1)=
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If $$x=\frac{7}{9}-\frac{15}{18}+\frac{10}{12} then (1-x)^{2}=$$
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If f(x) = x^3 - 5 and f(k) = 3 then k =
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For all numbers x and y, the operation фis defined by x ф y = (x+y)*(x-y) + (y-x)*(y+x) +x*y. What is the value of (sqrt12)ф(sqrt3)?
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For all positive numbers x, Δx is defined as the cube root of x, and ▽x is defined as the square root of x. If ▽(Δk) = m^2 , then k =
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sqrt( 81+81+81+81+81+81+81+81)=
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If f(x) = x^2 + 4 and f(2*k) = 36, then which of the following is one possible value of k?
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