Answer:
put a point on the 9 (the line on the left of 10) and on the -6 (the line to the left of -5).
a. 9 is located to the right of -6
b. use the symbol >
Step-by-step explanation:
Answer: The formula for the area A is
4t^2 + 56t + 160
Step-by-step explanation: Please see the attached diagram.
Figure ABCD shows the 20 cm by 8cm print. Then the surrounding frame which is shown in the diagram as EFGH has dimensions as t. That is, either side (length or width) measures t cm away from the edge of the picture frame. Therefore to compute the total area of the entire figure, which includes both picture and frame, we would have to add up t cm to the known dimensions. Hence the length of the entire figure would be 20 + t + t (which gives us 20 + 2t) and the width would be 8 + t + t (which would give us 8 + 2t).
Therefore if the length and width have been derived as above, the area of the entire figure can now be calculated as;
Area = (20 + 2t) x (8 + 2t)
By expanding the bracket we have
Area = 160 + 40t + 16t + 4t^2
By collecting like terms we now have
Area = 160 + 56t + 4t^2
Therefore
A = 4t^2 + 56t + 160
Answer:
$9
Step-by-step explanation:
The linear system is going to be:
16m + 13s + 29c = $419
26m + 19s + 45c = $649
6m + 11s + 0c = $96
I don't really know why it is bringing up the determinant, all you need to do is put this into reduced row echelon form. I personally used a calculator for this, but it could be done painstakingly by hand.
RREF gives me:
m=5
s=6
c=9
So a chocolate milkshake is $9