Answer:
tan−1(StartFraction 6.9 Over 9.8 EndFraction)
Step-by-step explanation:
tan−1(StartFraction 6.9 Over 9.8 EndFraction)
tan = opp/adj = 9.8/6.9
tan -1 = 1 / tan = 1 / (9.8 /6.9) = 6.9 /9.8
Answer: 2.76 g
Step-by-step explanation:
The formula to find the standard deviation:-
The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.
Then, 
Now, 
Then, 
Hence, the standard deviation of his measurements = 2.76 g
Answer:
x = -1
Step-by-step explanation:
4x + 5 = 3x + 4
-5 -5
4x = 3x -1
-3x = -3x
1x = -1
x = -1
Answer:
Step-by-step explanation:
6 and - 2 are the only two solutions to the required quadratic equation.
So, if the variable is represented by X then (X - 6) and (X + 2) will be the only two factors of the polynomial function.
Therefore, the equation is
(X - 6)(X + 2) = 0
⇒ 
If the leading coefficient of the equation is 3 then we can write the equation as (Answer)
Answer:
Step-by-step explanation:
Answer:
a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)
Explanation:
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and -8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
We integrate
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
