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Roman55 [17]
3 years ago
11

Please help asap!!! :(

Mathematics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

Our maximum is 19 when x=3 and y=7 and our minimum is -21 when x=3 and y = -3

Step-by-step explanation:

First, we can graph these inequalities out. As you can see in the picture, the three vertices where the inequalities all connect form a triangle. We can check each of these vertices to find our minimum and maximum.

First, we have (3,7). 4y-3x = 4(7)-3(3)=28-9=19

Next, for (3, -3), we have 4y-3x = 4(-3)-3(3) = -12-9=-21

Finally, for (0.5, 2), we have 4y-3x=4(2)-3(0.5)=8-1.5 = 6.5

Our maximum is 19 when x=3 and y=7 and our minimum is -21 when x=3 and y = -3

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3 years ago
Find the polynomial f(x) of degree 3 with real coefficients that has a y-intercept of 60 and zeros 3 and 1+3i.
Sauron [17]

\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0

[ correction added, Thanks to @stef68 ]

\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)

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3 years ago
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serious [3.7K]

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