I'm not all that familiar with that sort of representation, but I guess 3 3/4 = 15/4 and 2 1/6 = 13/6.
You will simply have to multiply the time spent walking, 13/6 hours, by the average distance traveled on an hourly basis, 15/4 km/h.
(13*15)/(4*6) = 195/24 = 8,125 km
Tom has 39 silver dollars. You can do this on your head. Jimmy has 32, and Tom 39.
If the sample is random, it is not valid.
Using relations in a right triangle, it is found that the length of AC is of 14 cm.
<h3>What are the relations in a right triangle?</h3>
The relations in a right triangle are given as follows:
- The sine of an angle is given by the length of the opposite side to the angle divided by the length of the hypotenuse.
- The cosine of an angle is given by the length of the adjacent side to the angle divided by the length of the hypotenuse.
- The tangent of an angle is given by the length of the opposite side to the angle divided by the length of the adjacent side to the angle.
Researching this problem on the internet, we have that:
- The opposite leg to angle A is of 48 cm.
Hence the hypotenuse is found as follows:
sin(A) = 48/h
0.96 = 48/h
h = 48/0.96
h = 50 cm.
The length of side AC is the other leg of the triangle, found using the Pythagorean Theorem, hence:
x = 14 cm.
More can be learned about relations in a right triangle at brainly.com/question/26396675
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Service charge for Julia at a beauty salon = $72.60
The rate of sales tax taken = 8%
Percentage of tips she added to the service charge before sales tax = 20%
Then amount of tips given by Julia = (20/100) * 72.60
= (1/5) * 72.60
= 14.52 dollars
Then the amount in total paid by Julia before adding service tax = (72.60 + 14.52) dollars
= 87.12 dollars
Now the sales tax of 8% has to be calculated on the amount of $87.12.
Then
Amount of sales tax given by Julia = (8/100) * 87.12
= 696.96/100
= 6.969
= 6.97 dollars
So the total amount paid by Julia for the service at the salon = (87.12 + 6.97) dollars
= 94.07 dollars
Julia paid a total of $94.07 for the service she received at the salon.