The answer is B) 4 : 16
Explanation:
RATIO FOR 1 MOLECULE OF ACETIC ACID:
2 carbon atoms : 2 oxygen atoms : 4 hydrogen atoms
Since its asking how many for 2 MOLECULES, we double the ATOMS (we multiply all by 2)
RATIO FOR 2 MOLECULES OF ACETIC ACID:
4 carbon atoms : 4 oxygen atoms : 8 hydrogen atoms
The question only asks for the ratio of carbon atoms to ALL atoms. To calculate all atoms, we add 4 + 4 + 8 = 16
So if there are 4 carbon atoms and the sum of all atoms is 16, out final ratio answer is:
B.) 4 : 16
Mark me brainliest please!!
Answer:
aver aver es 20 para arriba y 9 para abajo
1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m².
a = 51 m, b = 37 m, c = 20 m
semiperimeter: p = (51+37+20):2 = 54 m
Area of triangle:
![A=\sqrt{p(p-a)(p-b)(p-c)}\\\\A=\sqrt{54(54-51)(54-37)(54-20)}\\\\A=\sqrt{54\cdot3\cdot17\cdot34}\\\\A=\sqrt{9\cdot2\cdot3\cdot3\cdot17\cdot17\cdot2}\\\\A=3\cdot2\cdot3\cdot17\\\\A=306\,m^2](https://tex.z-dn.net/?f=A%3D%5Csqrt%7Bp%28p-a%29%28p-b%29%28p-c%29%7D%5C%5C%5C%5CA%3D%5Csqrt%7B54%2854-51%29%2854-37%29%2854-20%29%7D%5C%5C%5C%5CA%3D%5Csqrt%7B54%5Ccdot3%5Ccdot17%5Ccdot34%7D%5C%5C%5C%5CA%3D%5Csqrt%7B9%5Ccdot2%5Ccdot3%5Ccdot3%5Ccdot17%5Ccdot17%5Ccdot2%7D%5C%5C%5C%5CA%3D3%5Ccdot2%5Ccdot3%5Ccdot17%5C%5C%5C%5CA%3D306%5C%2Cm%5E2)
Rate: ₹ 3 per m².
Cost: ₹ 3•306 = ₹ 918
2. Find the area of the isosceles triangle whose perimeter is 11 cm and the base is 5 cm.
a = 5 cm
a+2b = 11 cm ⇒ 2b = 6 cm ⇒ b = 3 cm
p = 11:2 = 5.5
![A=\sqrt{5.5(5.5-3)^2(5.5-5)}\\\\ A=\sqrt{5.5\cdot(2.5)^2\cdot0.5}\\\\ A=\sqrt{11\cdot0.5\cdot(2.5)^2\cdot0.5}\\\\A=0.5\cdot2.5\cdot\sqrt{11}\\\\A=1.25\sqrt{11}\,cm^2\approx4.146\,cm^2](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B5.5%285.5-3%29%5E2%285.5-5%29%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B5.5%5Ccdot%282.5%29%5E2%5Ccdot0.5%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B11%5Ccdot0.5%5Ccdot%282.5%29%5E2%5Ccdot0.5%7D%5C%5C%5C%5CA%3D0.5%5Ccdot2.5%5Ccdot%5Csqrt%7B11%7D%5C%5C%5C%5CA%3D1.25%5Csqrt%7B11%7D%5C%2Ccm%5E2%5Capprox4.146%5C%2Ccm%5E2)
3. Find the area of the equilateral triangle whose each side is 8 cm.
a = b = c = 8 cm
p = (8•3):2 = 12 cm
![A=\sqrt{12(12-8)^3}\\\\ A=\sqrt{12\cdot4^3}\\\\ A=\sqrt{3\cdot4\cdot4\cdot4^2}\\\\A=4\cdot4\cdot\sqrt{3}\\\\A=16\sqrt3\ cm^2\approx27.713\ cm^2](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B12%2812-8%29%5E3%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B12%5Ccdot4%5E3%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B3%5Ccdot4%5Ccdot4%5Ccdot4%5E2%7D%5C%5C%5C%5CA%3D4%5Ccdot4%5Ccdot%5Csqrt%7B3%7D%5C%5C%5C%5CA%3D16%5Csqrt3%5C%20cm%5E2%5Capprox27.713%5C%20cm%5E2)
4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.
a = 2x
b = 3x
2x + 2•3x = 32 cm ⇒ 8x = 32 cm ⇒ x = 4 cm ⇒ a = 8 cm, b = 12 cm
p = 32:2 = 16 cm
![A=\sqrt{16(16-8)(16-12)^2}\\\\ A=\sqrt{16\cdot8\cdot4^2}\\\\ A=\sqrt{2\cdot8\cdot8\cdot4^2}\\\\ A=8\cdot4\cdot\sqrt2\\\\ A=32\sqrt2\ cm^2\approx45.2548\ cm^2](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B16%2816-8%29%2816-12%29%5E2%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B16%5Ccdot8%5Ccdot4%5E2%7D%5C%5C%5C%5C%20A%3D%5Csqrt%7B2%5Ccdot8%5Ccdot8%5Ccdot4%5E2%7D%5C%5C%5C%5C%20A%3D8%5Ccdot4%5Ccdot%5Csqrt2%5C%5C%5C%5C%20A%3D32%5Csqrt2%5C%20cm%5E2%5Capprox45.2548%5C%20cm%5E2)