Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 − 3 = x Square root of negative 2x plus 1 − 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 − 1 = x + 3 − 1 Square root of negative 2 x = x + 2 ( Square root of negative 2 x )2 = (x − 4)2 −2x = x2 − 8x + 16 −2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2)(x + 8) x + 2 = 0 x + 8 = 0 x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8 x = −2 x = −8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make?

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

pi x^2

step-by-step explanation:

pi times x squared

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5

step-by-step explanation:

its 512 because parenthesis and to the 2nd power is 512 so it will be 512