Yuto and Hila attempted to solve the same inequality. Their work is shown below.

Yuto’s Work

Hila’s Work

Negative two-thirds x + 9 greater-than 7. Negative two-thirds x + 9 minus 9 greater-than 7 minus 9. Negative two-thirds x greater-than negative 2. (Negative three-halves) (Negative two-thirds x) less-than (Negative three-halves) (Negative StartFraction 2 over 1 EndFraction). x less-than 3.

Negative two-thirds x + 9 greater-than 7. Negative two-thirds x + 9 minus 9 greater-than 7 minus 9. Negative two-thirds x greater-than 16. (Negative three-halves) (Negative two-thirds x) less-than (Negative three-halves) (StartFraction 16 over 1 EndFraction). x less-than negative 24.

Which statement is true about the work shown above?

Yuto is correct because he isolated the variable correctly and reversed the inequality symbol.
Yuto is incorrect because he should not have reversed the inequality symbol.
Hila is correct because she isolated the variable correctly and reversed the inequality symbol.
Hila is incorrect because she should not have reversed the inequality symbol.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

hope this !

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