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Setler79 [48]
3 years ago
14

Please use Distribution

Mathematics
1 answer:
Ronch [10]3 years ago
5 0
6*1/2 + 6*2/3 + 6*3/4
3 + 4 + 4.5
11.5
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Mr.Pope is buying Mrs. Pope's flowers. He buys 6 flowers for 10.95 without tax.
Montano1993 [528]
C

He payed $10.95 for 6 flowers. You want to find the price per each flower.

An equation to model this situation would be 6x = 10.95 where x = price per flower.

To solve this divide each side by 6.
10.95/6 = 1.825

Essentially to find the price per an individual thing you need to divide the total price by the amount of the item purchased.

He payed $1.825 per flower.
6 0
3 years ago
Read 2 more answers
-Translate this phrase into an algebraic expression.
Virty [35]

Answer:

20n - 4

Step-by-step explanation:

the product of 20 and a number means multiplying 20 by n (the variable representing the number)

to get four less than this product, you subtract 4

5 0
2 years ago
I Need The Answer Plz!!
Ivan

Answer:

DE

Step-by-step explanation:

See how DE cuts through half? We shouldn't be too accurate for the midsegment.

5 0
3 years ago
PLEASE I NEED THIS QUICKLY!!! DUE IN 30 MIN!!!
Dmitrij [34]

Answer:

171 times

Step-by-step explanation:

171 is the least common factor between numbers 9 and 19

6 0
3 years ago
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

6 0
3 years ago
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