Good evening ,
______
Answer:
4x² + 16x = 4×(x)×(x) + 4×(4)×(x)
= 4x(x+4)
:)
Given the basis β={(1,−1,3),(−3,4,9),(2,−2,4)}β={(1,−1,3),(−3,4,9),(2,−2,4)} and x=(8,−9,6)x=(8,−9,6), I am to find the corresponding coordinate vector [x]β[x]β. I claim that the coordinate vectors entries x1,x2,x3x1,x2,x3 meet the following criterion:
x1(1,−1,3)+x2(−3,4,9)+x3(2,−2,4)=(8,−9,6)x1(1,−1,3)+x2(−3,4,9)+x3(2,−2,4)=(8,−9,6)This is equivalent to solving the augmented matrix
⎡⎣⎢1−13−3492−248−96⎤⎦⎥[1−328−14−2−93946]which is row equivalent to
⎡⎣⎢100−31020−18−10⎤⎦⎥
2/7 x 14q+3 1/2 negative 3q=9
Answer 57q-3
1 / 5 = 0.2 pesos.
125 * 0.2 = 25 pesos el litro
The equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x
<h3>How to evaluate the expression?</h3>
The expression is given as:
(8x)^-2/3 * (27x)^-1/3
Evaluate the exponent 8^-2/3
(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * (27x)^-1/3
Evaluate the exponent (27x)^-1/3
(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * 1/3(x)^-1/3
Multiply 1/4 and 1/3
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^-2/3 * (x)^-1/3
Evaluate the exponent
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-2/3 -1/3)
This gives
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-1)
So, we have
(8x)^-2/3 * (27x)^-1/3 = 1/12x
Hence, the equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x
Read more about equivalent expression at
brainly.com/question/2972832
#SPJ1