Scott likes to run long distances. He can run 20 \text{ km}20 km20, start text, space, k, m, end text in 858585 minutes. He wants to know how many minutes (m)(m)left parenthesis, m, right parenthesis it will take him to run 52 \text{ km}52 km52, start text, space, k, m, end text at the same pace. How long will it take Scott to run 52 \text{ km}52 km52, start text, space, k, m, end text?

Given:

Scott can run 20 km in 85 minutes.

To find:

How long will it take Scott to run 52 km?

Solution:

Let us consider x be the number of minutes it will take him to run 52 km.

We know that,

By using this formula, we get

          ...(i)

            ...(ii)

On equating (i) and (ii), we get

On dividing both sides by 20, we get

Hence, it will take 221 minutes to run 52 km.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841