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Answer:
Latoya will charge $10 more to tutor for 5.5 hours
Step-by-step explanation:
We shall look at the total charges of both tutors individually.
Oscar charges $25 for the first hour of tutoring and $15 for each additional hour of fraction of an hour counting as a whole hour.
This means that for the first hour, Oscar charges $25.
For every subsequent hour, regardless of whether the student was tutored for the full hour or only partial, Oscar still charges $15 for the particular hour. To clarify, here is an example.
If a student is tutored for 2 hours, the cost for Oscar's tutoring is $25 (first hour) + $15 (second hour) = $40. However, if the student is still tutored for 1.5 hours, the cost for Oscar's tutoring is still $25 (first hour) + $15 (second hour) = $40. This is because Oscar counts the partial minutes as a full hour in tutoring costs.
For 5.5 hours, the cost of Oscar's tutoring is
$25 (first hour) + $15*4 (2nd to 5th hours) + $15 (the last half an hour) = $100.
Now, we shall look at Latoya's tutoring cost.
f(x) =
30x,
20x,
10x,
The tutoring cost depends on x, the number of tutoring hours. In this case, x is 5.5. The cost for 5.5 hour depends on the formula 20x since 5.5 is more than 3 but less than 6.
Hence, the tutoring cost for Latoya is $20*5.5 = $110.
In this case, Latoya charges $110-$100=$10 more to tutor for 5.5 hours.
Answer:
36.24 cm²
Step-by-step explanation:
The shaded region is the inscribed circle in the isosceles triangle. In order to find its area, we need to know the radius of the circle.
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One way to find the radius of the circle is to make use of the Angle Bisector theorem. It tells us an angle bisector divides the sides of a triangle proportionately. In the attached figure, that means angle bisector CD divides segment AE in proportion to sides CA and CE.
<h3>Side Length</h3>To make use of this theorem, we need to know the length of side CA. That is the hypotenuse of right triangle AEC, so can be found using the Pythagorean theorem. Isosceles triangle ABC is given as having height AE=10 and base BC = 12.
CA² = CE² +AE²
CA² = 6² +10² = 136 . . . . . . . CE is half the base length: 12/2 = 6
CA = √136 = 2√34
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<h3>Radius</h3>Now we can write the proportion using the angle bisector theorem.
DE/CE = DA/CA
r/6 = (10 -r)/(2√34)
r√34 = 3(10 -r) . . . . . . . multiply by 6√34
r(3 +√34) = 30
r = 30/(3 +√34)
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<h3>Area</h3>From this, we can find the area of the shaded circle to be ...
A = πr² = (3.14)(30/(3 +√34))²
= 3.14·900/(43 +6√34) ≈ 36.2374 . . . cm²
The area of the shaded region is about 36.24 cm².
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<em>Additional comment</em>
Using area considerations, the formula for the radius of the inscribed circle in an isosceles triangle with base 'b' and sides 'a' can be found to be ...
For a=√136 and b=12, this becomes ...
This is the same value we found above.
