Answer:
The equation for the distance Jane's trainer bikes is 
.
Step-by-step explanation:
We have attached diagram for your reference.
Given:
Distance traveled on bike towards south = 16 miles
Distance she ran towards west = 12 miles
We need to find distance Jane's trainer bikes.
Solution:
Let the distance Jane's trainer bike be 'x'.
Now we will assume it to be right angled triangle.
So by Pythagoras theorem which states that;
"Square of the third side is equal to the sum of square of the other two sides."
framing in equation form we get;
Hence the equation for the distance Jane's trainer bikes is .
On solving we get;
Hence Jane's trainer bikes a distance of 20 miles.
Amounttotal=amount per day times number of days
200=17.1 times d
divide both sides by 17.1
11.695=d
about 12 days
Slope = m
Equation to find slope is:
m = (y2 - y1) / (x2 - x1)
2 points:
(4,-2) 4 = x1 and -2 = y1
(8,5) 8 = x2 and 5 = y2
So...
m = (y2 - y1) / (x2 - x1)
m = (5 - (-2)) / (8 - 4)
m = 7 / 4 or 1.75
Therefore, the slope of the line is 7/4 OR 1.75.
L*w=area
(3x+21)*w=3x^2+33x+84
w=(3x^2+33x+84)/3x+21
(3(x+4)(x+7))/3(x+7)
w=x+4
<u>EXPLANATION</u><u>:</u>
Given that
sin θ = 1/2
We know that
sin 3θ = 3 sin θ - 4 sin³ θ
⇛sin 3θ = 3(1/2)-4(1/2)³
⇛sin 3θ = (3/2)-4(1/8)
⇛sin 3θ = (3/2)-(4/8)
⇛sin 3θ = (3/2)-(1/2)
⇛sin 3θ = (3-1)/2
⇛sin 3θ = 2/2
⇛sin 3θ = 1
and
cos 2θ = cos² θ - sin² θ
⇛cos 2θ = 1 - sin² θ - sin² θ
⇛cos 2θ = 1 - 2 sin² θ
Now,
cos 2θ = 1-2(1/2)²
⇛cos 2θ = 1-2(1/4)
⇛cos 2θ = 1-(2/4)
⇛cos 2θ = 1-(1/2)
⇛cos 2θ = (2-1)/2
⇛cos 2θ = 1/2
Now,
The value of sin 3θ /(1+cos 2θ
⇛1/{1+(1/2)}
⇛1/{(2+1)/2}
⇛1/(3/2)
⇛1×(2/3)
⇛(1×2)/3
⇛2/3
<u>Answer</u> : Hence, the req value of sin 3θ /(1+cos 2θ) is 2/3.
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