The graph represents part of brooke's trip to work. during what time period is her speed decreasing? a) between 0 and 10 minutes b) between 20 and 25 minutes c) between 10 and 15 minutes d) between 15 and 20 minutes

B) between 20 and 25 minutes

Step-by-step explanation:

First, note that the line from 0 minutes to 5 minutes increases 40 mph. (+40)

Next, note that the line from 5 minutes to 10 minutes increases 10 mph. (+10)

Next, note that the line continues to stay the same from 10 minutes to 20 minutes. (+0)

Now, we get to the last part, which shows a drop from 50 mph to 40 mph (50 - 40 = 10; -10 mph drop)

We can note that her speed starts at 40 mph, but drops to 10 mph at 5 - 10 minutes. However, note that there is no answer choice for it, and so the next logical choice is the drop of speed from 20 - 25, which is a 10 mph drop.

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

what is the model?

step-by-step explanation: