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.
Answer:
Option C (72)
Step-by-step explanation:
In the figure attached we can see a rectangle ABCD with length 12 cm and width 9 cm.
A triangle PEF with height BE and base EF.
Since side AE = 3 cm, so side EB = EC = 9 - 3 = 6cm
and EF = AD = 12 cm
Now we will find the area of of the parts of the card that are not shaded.
Area of parts that are not shaded = Area of rectangle - area of shaded triangle
= (12×9) -
= 108 -
Option c) 72 cm² is the area of non shaded part.
Hope this helps!! -Mina
Divide: 9x 2/36 - 4y 2/36 = 36/36. Simplify: x2/4 - y2/9 = 1. Then do (x-0) 2/4 - (y-0)2 =1. Hyperbola.
3/4 of 6/7. The of means multiplication.
= 3/4 * 6/7. 2 into 4 is 2 and 2 into 6 is 3.
= 3/2 * 3/7 No other number can divide, so we multiply out.
=9/14
That's the answer.
Answer:
A
Step-by-step explanation:
when you plug in 2,11
11 = 3*2 +5 that is correct
13 = 3*3 +5 that is not correct
only 2,11 works
