w= √7w
(2√5+3√7)^2
x= √(-3x+40)
(ab^3)(a^0b^ -2)
(4√3)/√16b
-6√5X^3 x √10x
(2√5+3√7)^2
x= √(-3x+40)
(ab^3)(a^0b^ -2)
(4√3)/√16b
-6√5X^3 x √10x
-
w = sqrt(7) * w
Well, what multiplied by anything equals itself?
(2*sqrt(5) + 3*sqrt(7))^2
This can be treated like:
(2*sqrt(5) + 3*sqrt(7)) * (2*sqrt(5) + 3*sqrt(7))
Then use FOIL to distribute.
2*sqrt(5) * 2*sqrt(5) + 2*sqrt(5) * 3*sqrt(7) + 3*sqrt(7) * 2*sqrt(5) + 3*sqrt(7) * 3*sqrt(7)
Then it's order of operations to simplify.
2*2*1 + 2*3*sqrt(5*7) + 2*3*sqrt(5*7) + 3*3*1
4 + 6*sqrt(35) + 6*sqrt(35) + 9
13+12*sqrt(35)
x = sqrt(-3x+40)
Square both sides.
x^2 = -3x + 40
x^2 - 40 = -3x
x^2 + 3x - 40 = 0
Can you get it from there? (hint: quadratic formula. or maybe factoring, don't remember how to tell.)
(a * b^3) * (a^0 * b^-2)
Associative Property of Multiplication
a * a^0 * b^3 * b^-2
When multiplying exponents of the same base, add together their powers.
a^(1+0) * b^(3-2)
a^1 * b^1
ab
(Side note on that last problem. If a * a^0 = a, what is a^0? Only one number does that! Any base raised to the 0th power is the number with that property.)
(4 * sqrt(3)) / sqrt(16b)
Can you confirm that's the problem? Is the b under the radical?
-6 * sqrt(5) * x^3...wait, I don't even know. Is that x multiplication, or actually an x? And are the x's under the radicals?
Well, what multiplied by anything equals itself?
(2*sqrt(5) + 3*sqrt(7))^2
This can be treated like:
(2*sqrt(5) + 3*sqrt(7)) * (2*sqrt(5) + 3*sqrt(7))
Then use FOIL to distribute.
2*sqrt(5) * 2*sqrt(5) + 2*sqrt(5) * 3*sqrt(7) + 3*sqrt(7) * 2*sqrt(5) + 3*sqrt(7) * 3*sqrt(7)
Then it's order of operations to simplify.
2*2*1 + 2*3*sqrt(5*7) + 2*3*sqrt(5*7) + 3*3*1
4 + 6*sqrt(35) + 6*sqrt(35) + 9
13+12*sqrt(35)
x = sqrt(-3x+40)
Square both sides.
x^2 = -3x + 40
x^2 - 40 = -3x
x^2 + 3x - 40 = 0
Can you get it from there? (hint: quadratic formula. or maybe factoring, don't remember how to tell.)
(a * b^3) * (a^0 * b^-2)
Associative Property of Multiplication
a * a^0 * b^3 * b^-2
When multiplying exponents of the same base, add together their powers.
a^(1+0) * b^(3-2)
a^1 * b^1
ab
(Side note on that last problem. If a * a^0 = a, what is a^0? Only one number does that! Any base raised to the 0th power is the number with that property.)
(4 * sqrt(3)) / sqrt(16b)
Can you confirm that's the problem? Is the b under the radical?
-6 * sqrt(5) * x^3...wait, I don't even know. Is that x multiplication, or actually an x? And are the x's under the radicals?
-
Damn it how could I forget ****...****...**** and now it's the summer