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3 years ago

Solve the inequality: -1/4(x^2-4x+3)>0

1 answer:

Montano1993 [528]3 years ago

5 0

Ok so assume that if you have
xy=0 then x and y=0
-1/4(x^2-4x+3)>0
multiply both sides by -4 to clear fraction (-4/-4=1)
flip sign
x^2-4x+3<0
factor
(x-3)(x-1)<0

we know that (+) times (-)=(-) and
(-) times (-)=(+) so we don't want them to be both negative, we want different sign
we cannnot have 3 since it would be (0)(2)<0 which is false
we cannot have 1 either since (-2)(0)<0 is also false
lets see if the solution is in betwen 3 and 1 or outside of 3 and 1
ltry 2
2^3-4(2)+3<0
8-8+3<0
3<0
false
therefor it is outside ie
x>3 and x<1

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Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

$\large\boxed{\large\boxed{161}}$

Explanation:

Let's put the information in a table step-by step.

(number of remaining students)

Juniors Seniors

Condition

Initially J S

15 seniors left S - 15

Twice juniors as seniors 2(S - 15)

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1/3 of seniors left 2/3×(S - 15)

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Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

$2/3\times (S - 15) - 1/4\times 2(S - 15) = 8$

Mutilply all by 12:

$8(S - 15)-6(S - 15)=96$

Distribution property:

$8S-120-6S-90=96$

Addtion property of equalities:

$8S-6S=96+120+90$

Add like terms:

$2S=306$

Division property of equalities:

$S=306/2=153$

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

$2/3\times (S-15)+1/4\times 2(S-15)$