Notice that
(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>
So we can add 1 to both sides of both equations, and we use the property above to get
<em>a</em> + <em>b</em> + <em>a b</em> = 76 ==> (1 + <em>a</em>)(1 + <em>b</em>) = 77
and
<em>c</em> + <em>d</em> + <em>c d</em> = 54 ==> (1 + <em>c</em>)(1 + <em>d</em>) = 55
Now, 77 = 7*11 and 55 = 5*11, so we get
<em>a</em> + 1 = 7 ==> <em>a</em> = 6
<em>b</em> + 1 = 11 ==> <em>b</em> = 10
(or the other way around, since the given relations are symmetric)
and
<em>c</em> + 1 = 5 ==> <em>c</em> = 4
<em>d</em> + 1 = 11 ==> <em>d</em> = 10
Now substitute these values into the desired quantity:
(<em>a</em> + <em>b</em> + <em>c</em> + <em>d</em>) <em>a</em> <em>b</em> <em>c</em> <em>d</em> = 72,000
Answer: The perimeter of a square is defined as the length that its boundary covers. The formula to calculate the perimeter of a square is given as: P = 4 × side.
Step-by-step explanation:
Look at the attached image (credit to Desmos graphing calculator)
To graph the equation y=4x+1, we just find two points that exist, plot them, and draw a straight line that goes through them.
Let's try x=0:
![y=4(0)+1=1](https://tex.z-dn.net/?f=y%3D4%280%29%2B1%3D1)
One point on the line is (0,1)
Let's try x=1:
![y=4(1)+1=5](https://tex.z-dn.net/?f=y%3D4%281%29%2B1%3D5)
Another point is (1,5).
Now, plot these two points and draw a line through them.