log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
|x|=√(x^2).
If 3x - 1 = 11 what is the value of x 2 + x?
(X - 2)2 = 2X -4
It is 2.
log33+log3x +2=3 log33+log3x=1 log3(3x)=1 3x=3 x=1 Other interpretation: log33+log3(x+2)=3 log3(3(x+2))=3 3(x+2)=27 x+2=9 x=7
log33+log3x +2=3 log33+log3x=1 log3(3x)=1 3x=3 x=1 Other interpretation: log33+log3(x+2)=3 log3(3(x+2))=3 3(x+2)=27 x+2=9 x=7
assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2
log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
Percent error is calculated using the formula: ((measured value - correct value) / correct value) x 100. Plugging in the values, we get ((3.24 - 3.02) / 3.02) x 100 = (0.22 / 3.02) x 100 ≈ 7.28%.
That depends what the value of x is.
2
2x3+nx2-x-4 Using the remainder theorem: n = -1 So the expression is now: 2x3-x2-x-4 When divided by x-2 = 2x2+3x+5 remainder 6 or 6 over x-2 The answer is correct because: (x-2)*(2x2+3x+5+6/x-2) = 2x3-x2-x-4
-if3X-1=11,what is the value of X^2+X?
2 x 2 x 2 x 2 x 5 = 80
|x|=√(x^2).
In order to write f(x) = |x| + |x-2| without the absolute value signs, it it necessary to write it as a piecewise function.We must define f as follows:f(x) = -2x + 2, if x < 0f(x) = 2, if 0