Let
x ----------> the height of the whole poster
<span>y ----------> the </span>width<span> of the whole poster
</span>
We need
to minimize the area A=x*y
we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2
so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4
find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]
for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in
(y)=[722/(x-4)]+2
y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in
the answer is
<span>the dimensions of the poster will be
</span>the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in
26x3=78
78-x=72
x=6
She needs to buy at least 6 more to have enough for 26 students
Answer:
LQ = 54
Median = 69
UQ = 94
Step-by-step explanation:
This list is already sorted for you, so you don't need to worry about that, otherwise you would need to sort the numbers in ascending order. To find the median, we do 
, where n is the amount of numbers. This gives us 4, so the median is at position 4, so the median is 69. The lower quartile is simply 
, so 2, so the lower quartile is 54. The upper quartile is 
, so 6, so the upper quartile is 94.
Answer:
-43
Step-by-step explanation:
1. -3 x -3 x -3 = -27
2. -27- 4 x 4
3. -27- 16 = -43
Answer:
the M is (2,5)
Step-by-step explanation:
