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xenn [34]
3 years ago
12

The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 21 s

eats and 30 wheels. The equation representing the total number of seats is u + b = 21 where u is the number of bicycles and b is the number of bicycles.
A. 2u + b = 21
B. u + 2b = 30
C. 2u + b = 30
D. u + 2b = 21
Mathematics
2 answers:
Sholpan [36]3 years ago
8 0

Answer:

The correct option is:

B. u + 2b = 30

Step-by-step explanation:

The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 21 seats and 30 wheels.

u is the number of unicycles and b is the number of bicycles.

There are total 21 seats

⇒  u+b=21

unicycle has 1 wheel and bicycle has 2 and total there are 30 wheels

⇒ u+2b=30

Hence, the correct option is:

B. u + 2b = 30

bearhunter [10]3 years ago
5 0
The wheels function is:

w=2b+u and w=30 so

u+2b=30


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We are given the following equations -

\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}  First, I swapped the first and third rows.

\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}  Leading coefficient of row 2 canceled.  

\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}  The start value of row 3 was canceled.

\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}       Matrix rows 2 and 3 were swapped.

\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}      Leading coefficient in row 3 was canceled.

\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}

The positioning of the zeros indicated that there was no solution!

<u><em>Hope that helps!</em></u>

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