Evaluate 4-0.25g+0.5h4−0.25g+0.5h4, minus, 0, point, 25, g, plus, 0, point, 5, h when g=10g=10g, equals, 10 and h=5h=5h, equals,
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I believe the correct given equation is in the form of:
4 – 0.25 g + 0.5 h
Now we are to evaluate the equation with the given values:
g = 10 and h = 5
What this actually means is that to evaluate simply means
to calculate for the value of the equation by plugging in the values of the
variables. Therefore:
4 – 0.25 g + 0.5 h = 4 – 0.25 (10) + 0.5 (5)
4 – 0.25 g + 0.5 h = 4 – 2.5 + 2.5
4 – 0.25 g + 0.5 h = 4
Therefore the value of the equation is:
4
To evaluate 17 int (sin^2 (x) cos^3(x))
From Trig identity. Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x)
Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x
So we have 17 int (sin^2x(cosx - cosxsin^2x))
int (sin^2x(cosx)dx - int (sin^4xcosx)dx. ----------(1)
Let u = sinx then du = cosxdx
Substituting into (1) we have
int (u^2du) - int (u^4du)
u^3/3 - u^5/5
Substitute value for u we have
(sinx)^3/3 - (sinx)^5/5
Hence we have 17 [ sin^3x/3 - sin^5x/5]
Answer:
12.083 units
Step-by-step explanation:
Here we are given with two coordinates and asked to determine the distance between them.
Here we are going to use the distance formula, which is given as under
Where
Replacing these values in the distance formula
Hence the distance is 12.083 units
<em>Greetings from Brasil...</em>
We need to use the Sine Law in Any Triangle....
(AB/SEN C) = (BC/SEN A)
19/SEN X = 16/SEN32
SEN X = 0,62
<em>using the sine arc</em>
ARC SEN 0,62 ≈ 39
