Answer:
Part 1) [tex]-3x^4+3x^3+8x^2-2x-5[/tex]
Part 2) The graph in the attached figure
Part 3) [tex]16y^8-24y^7+40y^5[/tex]
Part 4) [tex]x1=\frac{-1+\sqrt{139}i} {14}[/tex] and [tex]x2=\frac{-1-\sqrt{139}i} {14}[/tex]
Part 5) Harold reads more than [tex]90[/tex] pages, but less than [tex]180[/tex] page
Part 6) [tex](30,68)[/tex], [tex](40,60)[/tex]
Part 7) [tex]\$576[/tex]
Part 8) [tex]23 + 2i[/tex]
Step-by-step explanation:
The complete answers in the attached figure because is too long
Part 1) Subtract (enter answer is standard form)
we have
[tex](-2x^4-2x^3+8x^2+2)-(x^4-5x^3+2x+7)[/tex]
Eliminate parenthesis
[tex]-2x^4-2x^3+8x^2+2-x^4+5x^3-2x-7[/tex]
Combine like terms
[tex](-2x^4-x^4)+(-2x^3+5x^3)+8x^2-2x+(2-7)[/tex]
[tex]-3x^4+3x^3+8x^2-2x-5[/tex] ------> standard form
Part 2) Which graph represents the inequality?
we have
[tex]2x-3y\leq 6[/tex]
Rewrite the inequality ------> [tex]3y\geq2x-6[/tex]
the solution is the shaded area above the solid line
the equation of the line is equal to [tex]3y=2x-6[/tex]
the slope of the line is positive
the y-intercept of the line is the point [tex](0,-2)[/tex] ---> value of y when the value of x is equal to zero
the x-intercept of the line is the point [tex](3,0)[/tex] ---> value of x when the value of y is equal to zero
The solution in the attached figure
Part 3) Multiply
we have
[tex]8y^4(2y^4-3y^3+5y)\\=( 8y^4*2y^4)-(8y^4*3y^3)+(8y^4*5y)\\=16y^8-24y^7+40y^5[/tex]
Part 4) Solve for x
we have
[tex]0=7x^2+x+5[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]7x^2+x+5=0[/tex]
so
[tex]a=7\\b=1\\c=5[/tex]
substitute in the formula
[tex]x=\frac{-1(+/-)\sqrt{1^{2}-4(7)(5)}} {2(7)}[/tex]
[tex]x=\frac{-1(+/-)\sqrt{-139}} {14}[/tex]
remember that
[tex]i=\sqrt{-1}[/tex]
so
[tex]x=\frac{-1(+/-)\sqrt{139}i} {14}[/tex]
[tex]x1=\frac{-1+\sqrt{139}i} {14}[/tex]
[tex]x2=\frac{-1-\sqrt{139}i} {14}[/tex]
Part 5) Harold and Helen both spend [tex]3[/tex] hours reading a novel. Helen reads [tex]30[/tex] more pages than Harold. Helen reads at least [tex]40[/tex] pages in [tex]1[/tex] hour, but no more than [tex]70[/tex]pages in [tex]1[/tex] hour.
Which statement correctly describes the range of pages Harold reads?
Let
x------> the number of pages that Harold read
y------> the number of pages that Helen read
we know that
[tex]y=30+x[/tex] -----> equation A
[tex]y\geq 40\ pages[/tex] ------> in one hour
so
in three hours -------> [tex]y\geq 120\ pages[/tex] ------> inequality B
[tex]y\leq 70\ pages[/tex] ------> in one hour
so
in three hours ------->[tex]y\leq 210\ pages[/tex] ------> inequality C
Substitute equation A in the inequality B
[tex]x+30\geq 120\ pages[/tex] ------> [tex]x\geq 90\ pages[/tex]
Substitute equation A in the inequality C
[tex]x+30\leq 210\ pages[/tex] ------> [tex]x\leq 180\ pages[/tex]
therefore
Harold reads more than [tex]90[/tex] pages, but less than [tex]180[/tex] page
Part 6) Let
x------> the number of Standard Specials sold last Monday
y------> the number of Deluxe specials sold last Monday
we know that
[tex]2x+4y\geq 200[/tex] -----> inequality A
[tex]0.50x+1.25y\leq100[/tex] -----> inequality B
[tex]x\geq30[/tex] -----> inequality C
Remember that
If a ordered pair representing a combination of Standard specials and Deluxe specials could have been sold last Monday and make sense in the context of the situation
then
the ordered pair must satisfy all restrictions
Verify
A) [tex](30,68)[/tex]
Substitute the value of x and y in each inequality
Inequality A
[tex]2(30)+4(68)\geq 200[/tex]
[tex]332\geq 200[/tex] ------> is true
Inequality B
[tex]0.50(30)+1.25(68)\leq100[/tex]
[tex]100\leq100[/tex] -----> is true
Inequality C
[tex]30\geq30[/tex] -----> is true
The ordered pair [tex](30,68)[/tex] represent a combination of Standard specials and Deluxe specials could have been sold last Monday and make sense in the context of the situation
B) [tex](80,40.5)[/tex]
The value of [tex]y=40.5[/tex] not make sense in the context of the situation, because is not a whole number
C) [tex](60,70)[/tex]
Substitute the value of x and y in each inequality
Inequality A
[tex]2(60)+4(70)\geq 200[/tex]
[tex]400 \geq 200[/tex] ------> is true
Inequality B
[tex]0.50(60)+1.25(70)\leq100[/tex]
[tex]117.5\leq100[/tex] -----> is not true
therefore
the ordered pair [tex](60,70)[/tex] is not a solution
D) [tex](40,60)[/tex]
Substitute the value of x and y in each inequality
Inequality A
[tex]2(40)+4(60)\geq 200[/tex]
[tex]320\geq 200[/tex] ------> is true
Inequality B
[tex]0.50(40)+1.25(60)\leq100[/tex]
[tex]95\leq100[/tex] -----> is true
Inequality C
[tex]40\geq30[/tex] -----> is true
The ordered pair[tex](40,60)[/tex] represent a combination of Standard specials and Deluxe specials could have been sold last Monday and make sense in the context of the situation
E) [tex](50.5,40)[/tex]
The value of [tex]x=50.5[/tex] not make sense in the context of the situation, because is not a whole number