Utilizing the Law of Sines (sinA/a=sinB/b=sinC/c) and recognizing that the angles in a triangle add up to 180°.
The angle C calculates to 180-53-17=110°
Thus, we have 27/sin53=b/sin17=c/sin110
This leads to b=27sin17/sin53, c=27sin110/sin53
The perimeter is defined as a+b+c, so
p=27+27sin17/sin53+27sin110/sin53 units
p≈68.65 units (rounded to the nearest hundredth of a unit)
Response:
Here is the solution to the question:
Detailed Steps:
The following steps will be followed in this task:
- Step 1: Utilize the formulas tab on the sheet where you will apply the function found in the FLG "Function Library group".
- Step 2: Click on the Financial button, then select PMT.
- Step 3: After selecting PMT, input "B3/12" into the rate argument box.
- Step 4: In B4, fill in the Nper argument box with the value found in cell "B2" for the Pv argument box.
- Step 5: Click the OK button to finish.
The initial time was 15.7 minutes.
Explanation of steps:
Let x represent the initial time (in minutes).
The relay team is working on minimizing their relay race duration.
First, they cut down their time by 2.1 minutes.
Afterwards, they manage to reduce that time by 10 percent.
If the concluding time ends up being 3.96 minutes, then
x-t1-t2= 3.6
x= 3.6+ t1+ t2
x= 3.6+ 2.1+ 10
x= 15.7
Thus, their starting time was 15.7 minutes.
Response:
Step-by-step clarification:
Reply:
a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)
Clarification:
To address these equations, using direct integration is the simplest approach.
a) The equation provided is
dy / dt = -y + 8
dy / (y-8) = dt
We substitute variables
y-8 = u
dy = du
Substituting and integrating gives us
∫ du / u = ∫ dt
Ln (y-8) = t
Evaluating at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Simplifying the equation results in
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) The equation here is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
Integrating now
½ Ln (2y-5) = t
Evaluating at limits gives
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) The equation here bears a strong resemblance to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate this to get
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
