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Answer:
21 Pounds =
9.5254398 Kilograms
(rounded to 8 digits)
Step-by-step explanation:
- <em>Pounds : The pound or pound-mass (abbreviations: lb, lbm, lbm, ℔[1]) is a unit of mass with several definitions. Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces. </em>
- <em>Kilograms : The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Defined as being equal to the mass of the International Prototype Kilogram (IPK), that is almost exactly equal to the mass of one liter of water. The kilogram is the only SI base unit using an SI prefix ("kilo", symbol "k") as part of its name. The stability of kilogram is really important, for four of the seven fundamental units in the SI system are defined relative to it.</em>
Answer:
Step-by-step explanation:
The formula for simple interest is expressed as
I = PRT/100
Where
P represents the principal
R represents interest rate
T represents time in years
I = interest after t years
From the information given
T = 8 months = 8/12 = 2/3 years
P = $3000
R = 9.3%
Therefore
I = (3000 × 9.3 × 2/3)/100
I = 18600/100
I = $186
The maturity value (in dollars) of this loan would be
3000 + 186 = $3186
Answer:
<h2>4/3 Joules </h2>
Step-by-step explanation:
Work is said to be done when force applied to an object causes the object to move through a distance.
Work done = Force * perpendicular distance.
Given Force F = xy i + (y-x) j and a straight line (-1, -2) to (1, 2)
First we need to get the equation of the straight line given.
Given the slope intercept form y = mx+c
m is the slope
c is the intercept
m = y₂-y₁/x₂-x₁
m = 2-(-2)/1-(-1)
m = 4/2
m = 2
To get the slope we will substtutte any f the point and the slope into the formula y = mx+c
Using the point (1,2)
2 = 2+c
c = 0
y = 2x
Substituting y = 2x into the value of the force F = xy i + (y-x) j we will have;
F = x(2x) i + (2x - x) j
Using the coordinate (1, 2) as the value of s
![W = \int\limits^a_b ({2x^2 i + x j}) \, (i+2j)\\W = \int\limits^a_b ({2x^{2}+2x }) \, dx \\W = [\frac{2x^{3} }{3} +x^{2} ]\left \ x_2=1} \atop {x_1=-1}} \right.\\W = (2(1)^3/3 + 1^2) - (2(-1)^3/3 + (-1)^2)\\W =(2/3+1) - (-2/3+1)\\W = 2/3+2/3+1-1\\W = 4/3 Joules](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E2%20i%20%2B%20x%20j%7D%29%20%5C%2C%20%28i%2B2j%29%5C%5CW%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E%7B2%7D%2B2x%20%7D%29%20%5C%2C%20dx%20%5C%5CW%20%3D%20%5B%5Cfrac%7B2x%5E%7B3%7D%20%7D%7B3%7D%20%2Bx%5E%7B2%7D%20%5D%5Cleft%20%5C%20x_2%3D1%7D%20%5Catop%20%7Bx_1%3D-1%7D%7D%20%5Cright.%5C%5CW%20%3D%20%282%281%29%5E3%2F3%20%2B%201%5E2%29%20-%20%20%282%28-1%29%5E3%2F3%20%2B%20%28-1%29%5E2%29%5C%5CW%20%3D%282%2F3%2B1%29%20-%20%28-2%2F3%2B1%29%5C%5CW%20%3D%202%2F3%2B2%2F3%2B1-1%5C%5CW%20%3D%204%2F3%20Joules)
