The answer is b. The Remainder<span> Theorem says that if a polynomial </span>f(x<span>) is divided by </span>x<span> – k, then </span>the remainder<span> is </span>f(k). k=3 in our case, so the remainder is simply f(3). Plug in x=3 into <span>f(x) = 7x4 + 12x3 + 6x2 - 5x + 16. f(x)=946. So the answer is b.</span>
We can use linear combinations of the equations to eliminate variables.
3x - 4y = 1
-2x + 3y = 1
To eliminate y we'll make the linear combination of 3 times the first equation minus four times the second.
9x - 12y = 3
-8x + 12y = 4
Adding,
x = 7
We could solve for y directly but let's use another linear combination, twice the first plus three times the second:
2(3x - 4y) + 3(-2x + 3y)= 2(1)+3(1)
y = 5
Check: 3(7)-4(5)=1 good. -2(7)+3(5)=1 good.
Q18 Answer: (7,5)
y = -3x + 5
5x - 4y = -3
4y +1(5x - 4y) = 4(-3x + 5) + 1(-3)
5x = -12x + 20 -3
17 x = 17
x = 1
y = -3(1) + 5 = 2
Check: 5(1) - 4(2) = -3 good
Q19 Answer (1,2)
6x + 5y = 25
x = 2y + 24
6x = 12y + 144
5y = 25 - 12y - 144
17y = -119
y = -119/17= -7
x = 2y+24= 10
Check: 6(10)+5(-7)=25 good 2y+24=2(-7)+24=10=x good
Q20 Answer (10,-7)
3x + y = 18
-7x + 3y = -10
9x + 3y = 54
9x - -7x = 54 - -10
16x = 64
x=4
y = 18 -3x = 18-12=6
Check: 3(4)+6=18 good, -7(4)+3(6)=-10 good
Q21 Answer: (4,6)
$0.70 per candy.
$78 divided by 112 gives $0.696, so by rounding off, you get 0.70
Since 1A claims that the diagram is of a square, you can easily find the perimeter by multiplying just one side by 4, because the definition of a square says that all of its four sides are equal in length.
Take the left side, x and 4, and add them together, because both of these lengths add up to form the side of the square. You have found one side of the square, x + 4. Now multiply this side by 4 for the perimeter.
Perimeter is the length all around the figure, and since a square has 4 sides you would multiply one side by 4 to find the perimeter.
4(x + 4) is your expression for the perimeter of the square. You could probably solve 1B and 1C by substituting in 3 and 5 for x in the equation I've given you :)
