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allsm [11]
3 years ago
6

How do you solve this math problem of ratio?

Mathematics
1 answer:
Alex17521 [72]3 years ago
8 0
I don't know, this is tricky
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How find derivative of function: y=arcsin(2x+1)<br> ...?
Setler [38]
( \arcsin{x})'= \frac{1}{ \sqrt{1-x^2} } &#10;\\&#10;\\ (\arcsin{(2x+1)})'= \frac{1}{ \sqrt{1-(2x+1)^2} } (2x+1)'= \frac{2}{ \sqrt{1-(4x^2+4x+1)} } =&#10;\\&#10;\\= \frac{2}{ \sqrt{-4x^2-4x} } =\frac{2}{ \sqrt{4x(-x-1} )} =\frac{2}{ 2\sqrt{x(-x-1} )} =\frac{1}{ \sqrt{x(-x-1} )}
3 0
3 years ago
Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)
ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ],  slope is 86.3

111) for  [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope =  ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11)  [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111)  [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

3 0
3 years ago
50 point for this question
finlep [7]

Answer:

55.2° ≤ c ≤ 55.8°

Step-by-step explanation:

We know a + b + c = 180

Let's take the smallest value of a and b and find largest value of c:

51.3 + 72.9 + c = 180

c = 55.8

Now,

Let's take the largest value of a and b and find smallest value of c:

51.5 + 73.3 + c = 180

c = 55.2

Thus, we can say, range of c is:

55.2° ≤ c ≤ 55.8°

4 0
3 years ago
Read 2 more answers
Consider the following simplified algebraic expression.
Phoenix [80]

Answer:

*dies of boredom*

Step-by-step explanation:

EH, MWAH

7 0
3 years ago
How many quarters are there in three wholes?
Neporo4naja [7]

Answer:

1/4=1 whole so to keep it simple you can do 3x4=12and that how much quarters are in three whole

Step-by-step explanation:

6 0
3 years ago
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