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A clock is divided into 12 hours, meaning each hour is 30 degrees.

at 9:30, the minute hand will be at 6, and the hour hand will be halfway between 9 and 10. this will create an angle of 90 degrees (minute hand to 9 oclock) PLUS 15 degrees for the movement of the hour hand for a final answer of 105 degrees. 

Hope this helped!

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

\((45 min)(50 m/min) = 2250 m\)

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m\) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

\((50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875\)

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:51:52\)

Thanks! :)

So for this problem, let's use the arithmetic series formula.

We find the sum is \(\frac{(140+1)(70)}{2}\), which is \(141*35.\) Now that we have a factorization, we can easily find the greatest prime divisor.

The prime factorization is \(3*5*7*47\), so the greatest prime divisor is 47.

Thanks! :)

Let's know that XZW will form a triangle

\(\angle ZXW = 90^\circ \\ \angle XWZ = 60^\circ\)

Thus, \(\angle XZW = 180 - 90 - 60 = 30\)

30 degrees is our answer. 

Thanks! :)

An obtuse angle is an angle that measures in between \(90^\circ\) and \(180^\circ\)

Here's an image that lists all types of angles.

Notice that B is the only one that fits for the obtuse angle. 

A would be an acute angle since it is less than 90 degrees. 

C and D would be reflexive angles since they are more than 180 degrees, less than 360 degrees. 

The hour hand will move at a rate of 

\(360°  \space \text{in} \space 12 \space \text{hours} \\ 30° \space \text{in}  \space 1 \space \text{hour} \\ 30°/ 60  =  0.5° \space \text{in}  \text{ one minute}\)

So  at   4:15, the hour  hand has  moved   \(4 (30°) +  15(.5°)  =  [120+ 7.5]° = 127.5°\)from 12 o'clock

The minute hand  will move  at a rate of \(360°/ 60  = 6°\) per minute

So at 4:15 it will have moved  \([15* 6]°  = 90°\)  from the top of the hour

So...the smaller angle formed by the hands\(  = 127.5 - 90 = 37.5^\circ\)

May have made an error....

Thanks! :)

The area of a trapezoid is $80$. The length of one base is $16$ units greater than the other base, and one base of the trapezoid is $12$. Find the length of the median of the trapezoid.    

M = [ (12) + (12 + 16) ] / 2    

M = 40/2    

M = 20    

The median of a trapezoid is the average of its two bases.    

In this particular problem, the trapezoid's area is irrelevant.    

_.    

See image below:

Will rows for 45 minutes until Grace starts. In that time Will advances \(45 \cdot 50 = 2250\) meters. Every minute they get \(80\) meters closer to each other. At the time Grace starts, they will be \(550\) meters apart. It will take \(\frac{550}{80}\)minutes, which is around \(7\) minutes. They will meet at around \(\boxed{2:52} .\)

Draw a unit circle at the origin, O, and let P and Q loe pn this  circle

tan OP  = 4/1

cos OP = 1/sqrt(4^2 + 1^2)  = 1/sqrt 17      sin OP  = 4/sqrt 17

tan  OQ  = 5/1

cos OQ = 1/sqrt (5^2 + 1^2)  = 1/sqrt 26     sin QP = 5/sqrt  26

We can write P  as ( 1/sqrt 17 , 4/sqrt 17)

We can write Q as ( 1/sqrt 26 , 5/sqrt 26)

Slope  PQ  =

(5/sqrt 26 - 4/sqrt 17)  / ( 1/sqrt 26 - 1/sqrt 17)  =

(5sqrt 17- 4sqrt 26) / ( sqrt 17 - sqrt 26)  =

( 5 sqrt17 -4sqrt 26) (sqrt 17 + sqrt 26) / (17 -26)  =

(5*17 + sqrt 26 * sqrt 17 - 4*26) / -9

(-19 + sqrt 442) / -9 = 

(sqrt 442 - 19) / 9   ≈  -0.224

AB = 4  ???

https://web2.0calc.com/questions/geometry_50437

69 / 2553 = x / 1110

1110 * 69  = 2553 * x

x = 1110 * 69 / 2553   =  30

https://web2.0calc.com/questions/geometry_38859

Shortest altitude is drawn to the longest side

Area  =  sqrt [ 17 * (17 - 15) *(17 -10) * (17 - 9) ] = sqrt [17 * 2 * 7 * 8] = 4sqrt (119)

4sqrt (119)  = (1/2) 15 * shortest altitude

(8/15)sqrt (119) = shortest altitude ≈  5.82

See a similar problem here : 

https://web2.0calc.com/questions/trapezoid_85