Answer:
So Philip made 5 bracelets and 4 necklaces.
Step-by-step explanation:
Let x = number of bracelets and y = number of necklaces.
Since we have a total of 9 bracelets and necklaces,
x + y = 9 (1)
Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.
So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,
8x + 20y = 120 (2)
Simplifying it we have
2x + 5y = 30 (3).
Writing equations (1) and (3) in matrix form, we have
![\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%5C%5C30%5Cend%7Barray%7D%5Cright%5D)
Using Cramer's rule to solve for x and y,
![x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=x%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%261%5C%5C30%265%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)
x = (45 - 30) ÷ (5 - 2)
x = 15 ÷ 3
x = 5
![y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=y%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%269%5C%5C2%2630%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)
y = (30 - 18) ÷ (5 - 2)
y = 12 ÷ 3
y = 4
So Philip made 5 bracelets and 4 necklaces.
Step-by-step explanation:
D is the answer to your question
The rate of change in z at (4,9) as we change x but hold y fixed is =
3/[2sqrt(3x+2y)] put x = 4 , y = 9 = 3/[2sqrt(12+18) = 3/[2sqrt(30)] The
rate of change in z at (4,9) as we change y but hold x fixed is =
1/sqrt(3x+2y) put x = 4, y =9 = 1/sqrt(30)
Answer:
me please
Step-by-step explanation: