Considering the perimeter of the rectangle, we have that the length is of 9 inches and the width is of 55 inches.
<h3>What is the perimeter of a rectangle?</h3>
The perimeter of a rectangle of length l and width w is given as follows:
P = 2(l + w).
The length is an odd integer and the width is <u>5 times the next consecutive odd integer,</u> hence:
l = x, w = 5(x + 2).
The perimeter is of 128 inches, hence:
128 = 2l + 2w
128 = 2x + 10(x + 2)
128 = 2x + 10x + 20
12x = 108
x = 9.
Hence the length is of 9 inches and the width is of 55 inches.
More can be learned about the perimeter of a rectangle at brainly.com/question/10489198
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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.
Try photo math it is very helpful
Answer:
the correct answer is b :)))))!!!
The square root of 167 would be 12.922847. If you need it rounded to the nearest tenth it would be 12.9
If you need t rounded to the nearest hundredth, it would be 12.92.
Hope this helps :)