Answer:
see explanation
Step-by-step explanation:
(a)
← divide numerator and denominator by 4
=
= 0.8
(b)
→ divide numerator and denominator by 5
=
= 0.7
(c)
← divide numerator and denominator by 4
=
= 1.5
The perimeter of the square is:
ft
Step-by-step explanation:
The picture is attached with the answer.
the diagonal of a square divides it into two right angled-triangles.
We can use one of the triangle to find the length of side as the diagonal will be the hypotenuse of the triangle
Let a be the side of the square, then according to Pythagoras theorem
![d^2=a^2+a^2\\(10)^2 = 2a^2\\2a^2=100](https://tex.z-dn.net/?f=d%5E2%3Da%5E2%2Ba%5E2%5C%5C%2810%29%5E2%20%3D%202a%5E2%5C%5C2a%5E2%3D100)
Dividing both sides by 2
![\frac{2a^2}{2}=\frac{100}{2}\\a^2=50](https://tex.z-dn.net/?f=%5Cfrac%7B2a%5E2%7D%7B2%7D%3D%5Cfrac%7B100%7D%7B2%7D%5C%5Ca%5E2%3D50)
Taking Square root on both sides
![\sqrt{a^2}=\sqrt{50}\\a=\sqrt{25*2}\\a=\sqrt{5^2 * 2}\\a=5\sqrt{2}\ ft](https://tex.z-dn.net/?f=%5Csqrt%7Ba%5E2%7D%3D%5Csqrt%7B50%7D%5C%5Ca%3D%5Csqrt%7B25%2A2%7D%5C%5Ca%3D%5Csqrt%7B5%5E2%20%2A%202%7D%5C%5Ca%3D5%5Csqrt%7B2%7D%5C%20ft)
Now,
![P = 4a\\P = 4 * 5\sqrt{2}\\=20\sqrt{2}\ ft](https://tex.z-dn.net/?f=P%20%3D%204a%5C%5CP%20%3D%204%20%2A%205%5Csqrt%7B2%7D%5C%5C%3D20%5Csqrt%7B2%7D%5C%20ft)
Hence,
The perimeter of the square is:
ft
Keywords: Perimeter, Pythagoras Theorem
Learn more about Square at:
#learnwithBrainly
![HEY THERE !!](https://tex.z-dn.net/?f=%3Cb%3E%3Ci%3EHEY%20THERE%20%21%21)
As we know that cotx= 1/tanx
cot36° = 1/tan36°
=>cot36°= 1/0.726543
=>so cot 36°= 1.376382
So correct answer is option D.
HOPR YOU LIKED.
We have the following general rule for exponents:
![a^n\cdot a^m=a^{n+m}](https://tex.z-dn.net/?f=a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D)
then, in this case we have the following:
![3^3\cdot3^{-2}=3^{3+(-2)}=3^{3-2}=3^1=3](https://tex.z-dn.net/?f=3%5E3%5Ccdot3%5E%7B-2%7D%3D3%5E%7B3%2B%28-2%29%7D%3D3%5E%7B3-2%7D%3D3%5E1%3D3)
therefore, the answer is 3
Answer:
it is correct they weigh the same
Step-by-step explanation: