<h2>Answer :-</h2>
As we know that,
Pythagoras triplet
1) a² + b² = c²
Let
<h3>Hence, A can't be Pythagoras triplet</h3>
2) a² + b² = c²
<h3>Therefore, B can be Pythagoras triplet</h3>
3)a² + b² = c²
<h3>Hence, C can't be Pythagoras triplet</h3>
4) a² + b² = c²
<h3>Hence, D can't be Pythagoras triplet</h3>
<h2 /><h2>Therefore :-</h2>
Only B can be Pythagoras triplet.
Answer:
m=11÷600
Step-by-step explanation:
150m-100m+47.750=50.5000-200m
collect the like terms
-50m+47.75=50.5-200m
move variable to the left side and change its sign
-50m+200m+47.75=50.5
Move constant to the right side and change its sign
50m+200m=50.5-47.75
collect the like terms(again)
150m=50.5-47.75
subtract the numbers
150m=2.75
divied both sides of the equation by 150
m=11÷600
solution
m=11÷600
alternative form
m=0.0183
1. {0, all EVEN numbers, all ODD numbers|
2. {0}
3.{0, all EVEN numbers, 2^2, 3^2, 4^2, ,5^2, 6^2)
4.{2^2, 4^2,6^2}
5. 2^4
6.1/16 x^-3 y^-9 z^4 = z^4 / 16x^3y^9
7. x^2y^2z^5
8. 1.35 x 10^-2
9. 4 x10^-2
Answer:
ok sure
Step-by-step explanation:
Let's look at the picture, let's imagine that the gray line is the perimeter fence and that the red OR the blue is the one dividing it. We can see that the blue line is longer than the red one, so it will be advantageous, to have a bigger area, to have the dividing fence the smallest possible.
Let's say then that the width (W) is bigger (or equal) to the length (L), so we have:
The area is W*L, so we have
this function is a parabola facing down, its zeros are 0 and 80, therefore its maximum is when L=40
hence, L=40 and W=(240-120)/2=60
It will be a rectangle, measuring 60x40 and the divinding fence will be 40