Answer:
The dimensions of the rectangular poster is 15 in by 5 in.
Step-by-step explanation:
Given that, the area of the rectangular poster is 75 in².
Let the length of the rectangular poster be x and the width of the rectangular poster be y.
The area of the poster = xy in².

....(1)
1 in margin at each sides and 3 in margin at top and bottom.
Then the length of printing space is= (x-2.3) in
=(x-6) in
The width of printing space is = (y-2.1) in
=(y-2) in
The area of the printing space is A =(x-6)(y-2) in²
∴ A =(x-6)(y-2)
Putting the value of y


Differentiating with respect to x

Again differentiating with respect to x

To find the minimum area of printing space, we set A' = 0




Now putting x=±15 in A''

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.
From (1) we get

Putting the value of x

=5 in
The dimensions of the rectangular poster is 15 in by 5 in.
Given:
A(3,0)
B(1,-2)
C(3,-5)
D(7,-1)
1) reflect across x=-4
essentially calculate the difference between the x=-4 line and Px and "add" it in the other direction to x=-4
A(-4-(3-(-4)),0)=A(-11,0)
B(-4-(1-(-4)),-2)=B(-9,-2)
C(-4-(3-(-4),-5))=C(11,-5)
D(-4-(7-(-4)),-1)=D(-15,-1)
2) translate (x,y)->(x-6,y+8)
A(-3,8)
B(-5,6)
C(-3,3)
D(1,7)
3) clockwise 90° rotation around (0,0), flip the x&y coordinates and then decide the signs they should have based on the quadrant they should be in
A(0,-3)
B(-2,-1)
C(-5,-3)
D(-1,-7)
D) Dilation at (0,0) with scale 2/3, essentially multiply all coordinates with the scale, the simple case of dilation, because the center point is at the origin (0,0)
A((2/3)*3,(2/3)*0)=A(2,0)
B((2/3)*1,(2/3)*-2)=B(2/3,-4/3)
C((2/3)*3,(2/3)*-5)=C(2,-10/3)
D((2/3)*7,(2/3)*-1)=D(14/3,-2/3)
Answer:
linear model
Step-by-step explanation:
linear graph will produce a straight line
Answer:
The answer would be 15 cookies.
Step-by-step explanation:
Cookies = 24 cents each
Pop = 45 cents each