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What is the answer to -2 equals 4r plus s solving for s?
-2=4r+s s=-4r-2 or s=-(4r+2)
What is 2q divided by q?
2q ÷ q = 2
What is 4r r in simplest form?
4r + 4 = 5r 4r - r = 3r 4r x r = 4r^2 4r/r = 4
What is the product 2 of q?
2q
Completely factor the expression 4q2 27r4?
To factor the expression 4q^2 - 27r^4, we first note that this is a difference of squares, as 4q^2 is (2q)^2 and 27r^4 is (3r^2)^3. Therefore, we can rewrite the expression as (2q)^2 - (3r^2)^2. This can be factored further using the difference of squares formula, resulting in (2q + 3r^2)(2q - 3r^2) as the completely factored form.
How do you simply 4r r?
To simplify the expression (4r \cdot r), you multiply the coefficients and combine the like terms. This results in (4r^2). Thus, the simplified form of (4r \cdot r) is (4r^2).
Use the identity below to complete the tasks:a3 + b3 = (a + b)(a2 - ab + b2)Use the identity for the sum of two cubes to factor 8q6r3 + 27s6t3.What is a What is b?
Below ^ denotes power. 8q^6r^3=(2q^2r)^3 denote as a^3 and find a=2q^2r 27s^6t^3=(3s^2t)^3 denote as b^3 and find b=3s^2t Now 8q^6r^3+27s^6t^3=a^3+b^3 = (a+b)(a^2-ab+b^2) | substitute back =(2q^2r+3s^2t)(4q^4r^2-6q^2rs^2t+9s^4t^2). Notice (2q^2r)^2=4q^4r^2 (3s^2t)^2=9s^4t^2. Hence 8q^6r^3+27s^6t^3=(2rq^2+3ts^2)(4r^2q^4-6rts^2t^2+9t^2s^4), a=2q^2r and b=3s^2t.
What is the answer to -2 equals 4r plus s solving for s?
-2=4r+s s=-4r-2 or s=-(4r+2)
How would you simplify this expression 5p times 2q divided by 4p?
5p x 2p = 10p, then 10p / 4p = 2 remainder 2
What is 2q divided by q?
2q ÷ q = 2
What is 4r r in simplest form?
4r + 4 = 5r 4r - r = 3r 4r x r = 4r^2 4r/r = 4
Why is sqrt 2 irrational?
First we will assume that sqrt(2) is rational, meaning that it can be written as a ratio of two integers say (p/q) p and q must have no common factors sqrt(2)=p/q, square both sides 2=p^2/q^2, multiply both sides by q^2 2q^2=p^2, since 2 divides by the LHS, so does the RHS, meaning that p^2 is evenand because p^2 is even, so is p itselfLet p=2r with r being an integer so that p^2=2q^2=(2r)^2=4r^ 2Since 2q^2=4r^2, q^2=2r^2Because q^2 is 2r^2, then q^2 is even, meaning q itself is evenSince p and q are even, they have a common factor of 2THEREFORE, sqrt(2) cannot be rational
How do you get -2 equals 4r plus s solving for s?
s = -2 -4r
What is the product 2 of q?
2q
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q on the Cartesian plane?
Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0
How do you solve this problem 2q equals -14?
16
How do I justify with steps 8(2q)?
8(2q) = (8*2)q = 16*q = 16q.
