Your answer is 9s^2-9st+6st-6t^2
Remark
This is quite a nice little problem. It takes a minute or three to figure out the answer, and when you do, you will be certain that you have been tricked. It is a little like the egg of Columbus.
Solution
The Base of Triangle ABN is AB
The Base of Triangle CDM is CD
The height of both given triangles is h. That is the distance between the two parallel lines.
Area ABN = 1/2*AB * h = 23 cm^2
Area CDM = 1/2*CD * h = 18 cm^2
Now the Area of the trapezoid is
Area_Trapezoid = 1/2 * h (AB + CD) Using the distributive property Remove the brackets.
Area_Trapezoid = 1/2*AB*h + 1/2*CD*h Did you notice something? Those terms are just the area of the triangles (written above.)
Area Trapezoid = 23 + 18 = 41 cm^2 <<<< Answer
C(n)=2.00+3.20n
If n = 4 miles:
C(4)=2+3.20(4)
C(4)=2+12.80
C(4)=14.80
The cost of the taxi is $14.80.
Explanation:
Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that
a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.
Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.
Answer:
170 children
74 students
85 adults
Step-by-step explanation:
Given
Let:
For the capacity, we have:
For the tickets sold, we have:
Half as many as adults are children implies that:
Required
Solve for A, C and S
The equations to solve are:
-- (1)
-- (2)
-- (3)
Make C the subject in (3)
Substitute in (1) and (2)
-- (1)
Make S the subject
-- (2)
Substitute
Solve for A
Recall that:
Recall that:
Hence, the result is:
