Answer:
perimeter = 20.9 units
Step-by-step explanation:
perimeter
perimeter = distance around two dimensional shape
= addition of all sides lengths
<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>
distance formula:
<h3>1) distance of AB</h3>
A(-3,0) B(2,4)
x1 = -3 x2 = 2
y1 = 0 y2 = 4
(substitute the values into the distance formula)
AB = 6.4 units
<h3>2) distance of BC</h3>
B(2,4) C(3,1)
x1 = 2 x2 = 3
y1 = 4 y2 = 1
BC = 3.2 units
<h3>3) distance of CD</h3>
C(3,1) D(-4,-3)
x1 = 3 x2 = -4
y1 = 1 y2 = -3
CD = 8.1 units
<h3>4) distance of AD</h3>
A(-3,0) D(-4,-3)
x1 = -3 x2 = -4
y1 = 0 y2 = -3
AD = 3.2 units
<h2>perimeter of figure</h2>
= AB+BC+CD+AD
= 6.4 + 3.2 + 8.1 + 3.2
= 20.9 units
I believe 1/9=9, 2/9=18, 3/9=27, 4/9=36
Don't get mad if you get it wrong.
Answer:
Takeru bought 72 eggs and baked 18 soufflés .
Step-by-step explanation:
Let
e = number of eggs Takeru bought.
s = number of soufflés Takeru bought.
Then we know that Takeru bought 4 times as many eggs as he baked soufflés, therefore:
And since for every soufflés Takeru bakes he uses 3 eggs, and after backing he has 18 eggs left, we have:
<em>This says that from eggs there are 18 eggs left after Tkeru used three eggs for each soufflé.</em>
Now we have two equations:
We put the value of e from equation (1) into equation (2) and solve for s and get:
Thus
Number of eggs Takeru bought = 72.
Number of soufflés Takeru bought = 18.
Given 2.50x + 3.50y < 30.
Where x represent the number of hamburgers and y represent the number of cheeseburgers.
Now question is to find the maximum value of hamburgers Ben could have sold when he has sold 4 cheeseburgers.
So, first step is to plug in y=4 in the given inequality. So,
2.50x+3.50(4)<30
2.50x+14 <30
2.50x<30- 14 Subtracting 14 from each sides.
2.50x< 16
Dividing each sides by 2.50.
x<6.4
Now x being number of hamburgers must be an integer , so tha maximum value of x can be 6,
thus x = 6 hamburgers
So, the maximum value of hamburgers Ben could have sold is 6*2.5=$15
Hope this helps!!