Mathematics, 03.04.2020 21:28, username999990

PLEASE ANSWER! IT'S SOO HARD FOR ME!!

a. Surya is competing in a triathlon that includes swimming, bicycling and running. For the first two segments of the race, Surya swam at an average rate of 4 kilometers per hour and bicycled at an average rate of 40 kilometers per hour. It took 81.75 minutes for Surya to swim and cycle a combined total of 43.29 kilometers, completing the first two segments of the race. How long, in minutes, did it take him to complete the swimming portion of the race? Express your answer to the nearest hundredth.

b. If an athlete who completes this triathlon swims, bicycles and runs a combined total of 52.95 kilometers, the distance the athlete swam is what percent of the total race distance? Express your answer as a percent to the nearest hundredth.

c. At what average rate, in kilometers per hour, must Surya run the third, and final, portion of the race if he wishes to complete the entire race in no more than 140 minutes? Express your answer to the nearest hundredth.

Answers: 2

Show answers

Answers

answered: insulation62

a) 18.68 minutes

b) 2.34%

c) 9.96 km/h

Step-by-step explanation:

The velocity of Surya in the 1st segment (swimming part) is

v_1=4 km/h

So, the distance covered by Surya in the swimming part is

d_1=4t_1

where t_1 is the time taken to cover this part.

The velocity of Surya in the 2nd segment (biking part) is

v_2=40 km/h

So the distance covered in the biking part is

d_2=40t_2

where t_2 is the time taken to cover this part.

We know that the total distance covered in these 2 segments is

d=d_1+d_2=43.29

So it can be rewritten as

4t1+40t_2=43.29 (1)

The total time taken to cover these two parts is

t=t_1+t_2=81.75 min =1.36 h

So we have

t_2=1.36-t_1

And substituting in (1) we can find t1:

$4t_1 + 40(1.36-t_1)=43.29\\36t_1=11.21\\t_1=\frac{11.21}{36}=0.31 h = 18.68 min$

The total distance covered in all 3 segments is

d=d_1+d_2+d_3=52.95 km

where

d1 is the length of the swimming part

d2 is the length of the biking part

d3 is the length of the running part

We know that the distance covered in part 1 (swimming part) is

d_1=v_1 t_1 = (4)(0.31)=1.24 km

Therefore we can find the percent that the swimming part correspond to the total race as:

$\frac{d_1}{d}=\frac{1.24}{52.95}=0.0234 = 2.34\%$

Here the total time taken by Surya for the race must be

t=140 min = 2.33 h

And this is the sum of the times of the 3 segments:

t=t_1+t_2+t_3

where

t1 is the time taken to complete the swimming part

t2 is the time taken to complete the biking part

t3 is the time taken to complete the running part

We know already that

t_1=0.31 h

and

t_2=1.36-t_1=1.36-0.31=1.05 h

So the time for the 3rd section must be

t_3=t-t_1-t_2=2.33-0.31-1.05=0.97h

The distance to cover in the last part is

d_3=d-d_1-d_2=52.95-43.29=9.66 km

So the average velocity of the last segment must be:

$v_3=\frac{d_3}{t_3}=\frac{9.66}{0.97}=9.96 km/h$

answered: Guest

you would need to know all the numbers that are on there

step-by-step explanation:

answered: Guest

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12