Answer:
73920
Step-by-step explanation:
Number of ways to choose 9 appetizers from 12: ₁₂C₉
Number of ways to choose 3 main courses from 8: ₈C₃
Number of ways to choose 2 desserts from 4: ₄C₂
The total number of ways is:
₁₂C₉ × ₈C₃ × ₄C₂
= 220 × 56 × 6
= 73920
Answer:
The height of the tree is is 60m
Step-by-step explanation:
Let's answer a, as it is the only complete question.
We know that the angle of elevation of the top of a tree observed from a point 60m away, is 45°.
We can model this with a triangle rectangle, a sketch of it can be seen below (assuming that you are looking it from the ground).
You can see that the adjacent cathetus to the 45° angle is equal to 60m
And the opposite cathetus is the measure we want to find.
Now you can remember the trigonometric relation:
tan(a) = (opposite cathetus)/(adjacent cathetus).
So to find the height of the tree we need to solve:
tan(45°) = H/60m
This is just:
tan(45°)*60m = H =60m
The height of the tree is is 60m
<h2>
Points of Intersection</h2>
To find the point of intersection of two lines, we can use substitution to find the <em>x</em> and <em>y</em> coordinates.
<h2>Solving the Question</h2>
We're given:
Use substitution to find <em>x</em>:

Use substitution to find <em>y</em>:

<h2>Answer</h2>
The point of intersection is (2,-2).
Answer:
c. x ≈ 1.71, y ≈ 6.29, z ≈ 60.71
Step-by-step explanation:
You are at a bit of a disadvantage with this problem, because both the hint and the answer choices are incorrect.
The matrix equation is ...
![AX=B\\\\\left[\begin{array}{ccc}-1&1&0\\4&3&0\\1&-7&1\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right]=\left[\begin{array}{c}8&12&15\end{array}\right]](https://tex.z-dn.net/?f=AX%3DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%261%260%5C%5C4%263%260%5C%5C1%26-7%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2612%2615%5Cend%7Barray%7D%5Cright%5D)
Contrary to what the hint is telling you, the multiplication by the inverse must be on the left:

Using appropriate tools to compute the inverse, this becomes ...
![\left[\begin{array}{c}x&y&z\end{array}\right]=\dfrac{1}{7}\left[\begin{array}{ccc}-3&1&0\\4&1&0\\31&6&7\end{array}\right]\left[\begin{array}{c}8&12&15\end{array}\right]\\\\\left[\begin{array}{c}x&y&z\end{array}\right]=\dfrac{1}{7}\left[\begin{array}{c}-12&44&425\end{array}\right]\approx\left[\begin{array}{c}-1.714&6.286&60.714\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B7%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%261%260%5C%5C4%261%260%5C%5C31%266%267%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2612%2615%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B7%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-12%2644%26425%5Cend%7Barray%7D%5Cright%5D%5Capprox%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1.714%266.286%2660.714%5Cend%7Barray%7D%5Cright%5D)
The closest match is choice C.