Classify the following as an element, a compound, or a mixture: sucrose
A. Element
B. Compound
C. Mixture

Sucrose is B. Compound.

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Solve the following quadratic equation by completing the square: x2 + 5x - 3 = 0
A. (-5+v37)/2, (-5 -v37)/2
B. (5+v37)/2, (5 -v37)/2
C. (-5+v13)/2, (-5 -v13)/2
D. No real solution **Weegy:** x^(2)+5x-3=0
Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.
x^(2)+5x=3
To create a trinomial square on the left-hand side of the equation, [ add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add ((5)/(2))^(2) to both sides of the equation.
x^(2)+5x+(25)/(4)=3+(25)/(4)
To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is 4. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
x^(2)+5x+(25)/(4)=3*(4)/(4)+(25)/(4)
Complete the multiplication to produce a denominator of 4 in each expression.
x^(2)+5x+(25)/(4)=(12)/(4)+(25)/(4)
Combine the numerators of all fractions that have common denominators.
x^(2)+5x+(25)/(4)=(12+25)/(4)
Add 25 to 12 to get 37.
x^(2)+5x+(25)/(4)=(37)/(4)
Factor the perfect trinomial square into (x+(5)/(2))^(2).
(x+(5)/(2))^(2)=(37)/(4)
Take the square root of each side of the equation to setup the solution for x.
~((x+(5)/(2))^(2))=\~((37)/(4))
Remove the perfect root factor (x+(5)/(2)) under the radical to solve for x.
(x+(5)/(2))=\~((37)/(4))
Split the fraction inside the radical into a separate radical expression in the numerator and the denominator. A fraction of roots is equivalent to a root of the fraction.
(x+(5)/(2))=\(~(37))/(~(4))
Pull all perfect square roots out from under the radical. ] (More)

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