Answer:
x= -10/3
y= -8/9
Step-by-step explanation:
2x+1=3(y−1);6(2y−1)=5x
Rewrite equations:
2x+1=3y−3;12y−6=5x
Step: Solve2x+1=3y−3for x:
2x+1=3y−3
2x+1+−1=3y−3+−1(Add -1 to both sides)
2x=3y−4
2x2=3y−42(Divide both sides by 2)
x=32y−2
Step: Substitute32y−2forxin12y−6=5x:
12y−6=5x
12y−6=5(32y−2)
12y−6=152y−10(Simplify both sides of the equation)
12y−6+−152y=152y−10+−152y(Add (-15)/2y to both sides)
92y−6=−10
92y−6+6=−10+6(Add 6 to both sides)
92y=−4
92y92=−492(Divide both sides by 9/2)
y=−89
Step: Substitute−89foryinx=32y−2:
x=32y−2
x=32(−89)−2
x=−103(Simplify both sides of the equation)
A²+b²=c²
a=b because it has 2 45° angles and one 90 so it is isosceles right triangle
∴ a² + a² = c²
2a²=24²
2a²=576
a²=288
a=√288
a=16.97056275
Therefore. With A=1/2bh
A=(1/2) * 16.97056275 * 16.97056275
= (1/2) * 288
= 144
144 ft²
Answer:
We can use seven letters and numbers.
I am assuming that any numeral in the range 0..9 or any letter from the English alphabet A..Z can appear in any position, with no blank spaces allowed and no restrictions on repetition. I am also assuming that plates with fewer than seven letters and numbers are not allowed.
So, for example A879BX8 is acceptable, so are 5555555 and ABCDEFG, but not A.123.ZX or…..7A, where the dot represents a space.
I am also assuming that you can only use upper case letters.
With these restrictions, there are 36 possibilities for each space and the total number of valid number plates would be 36^7 = 78,364,164,096, let's say about 78 billion.
It is estimated that there are about 1.3 billion cars, trucks and buses in the road today. This number plate system therefore allows more than enough unique license plates. I'd even hazard a guess that it might be more than enough for every road vehicle that has ever been built or ever will be.
In practice there would be other restrictions, for example only letters in some positions and only numbers in others. There'd still be plenty to go around.
Step-by-step explanation:
1 2/3y = 3x
A picture is worth a thousand words
