Answer:
Step-by-step explanation:
To solve each of these all you have to do is multiply 60kg which is the whole mass of a human body times the percentage so..
Oxygen = (0.65)(60) = 39kg
Carbon = (0.18)(60) = 10.8kg
Hydrogen = (0.10)(60) = 6
Nitrogen = (0.03)(60) = 1.8kg
Calcium = (0.015)(60) = 0.9kg
Phosphorus = (0.01)(60) = 0.6kg
These numbers dont exactly add up to 60 because the percentages gives dont add up to 100%, but this is how to do it.
Answer:
your answer is 200 if you add 84 and 116 but is you add 25 82 213 and 97 it will be 417
Step-by-step explanation:
Answer:
127/12
Step-by-step explanation:
4 × 2 + 12x = 135
(1. Simplify 4 x 2 to 8.
8 + 12x = 135
(2. Subtract 88 from both sides.
12x= 135 - 8
(3. Simplify 135 - 8 to 127
12x = 127
(4. Divide both sides by 12
x= 127/12
Decimal Form: 10.583333
I think this is the awnser, but don't quote me on that
Answer:
The value of x is 11 degrees
Step-by-step explanation:
Here, given that two angles are vertical angles, we are interested in calculating the value of x. We simply want to use an important property of vertical angles to get the value of x as given in the question.
When we talk of vertical angles, they are angles which lies on opposite side to each other when we have two intersecting line.
By property, vertically opposite angles are equal.
Thus, in this case, we are to equate the angles we are having since they are both vertical angles;
Thus,
Mathematically;
angle A = angle B
3x + 3 = 4x -8
Collect like terms;
4x-3x = 3 + 8
x = 11 degrees
the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5CA%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5CqquadB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquadd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B%5B0-%28-2%29%5D%5E2%2B%5B5-%28-2%29%5D%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B%280%2B2%29%5E2%2B%285%2B2%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B4%2B49%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B53%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5C%5C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-0%29%5E2%2B%281-5%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B3%5E2%2B%28-4%29%5E2%7D)
![\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20BC%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20%5Cboxed%7BBC%3D5%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CC%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2-3%29%5E2%2B%28-2-1%29%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%29%5E2%7D%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B25%2B9%7D%5Cimplies%20%5Cboxed%7BCA%3D%5Csqrt%7B34%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C~%5Chfill%20%5Cstackrel%7BAB%2BBC%2BCA%7D%7B%5Capprox%2018.11%7D~%5Chfill)
