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bazaltina [42]
3 years ago
11

Ms. Ironperson and Mr. Thoro are making Avenger posters to give children when they visit Avenger Academy. Ms. Ironperson has com

pleted 12 posters and will complete 6 more per day. Mr. Thoro has not started yet but can make 12 per day. At some point Mr. Thoro will catch up and both will have finished the same number of posters. When this does happen, how many posters will each Avenger have completed? If x denotes the number of days and y denotes the number of posters, what are the equations needed to solve this problem?
Mathematics
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

1) When this does happen, how many posters will each Avenger have completed?

24 posters each.

2) If x denotes the number of days and y denotes the number of posters, what are the equations needed to solve this problem?

y = 12 + 6x.......... Equation 1

y = 12x .......... Equation 2

Step-by-step explanation:

1) When this does happen, how many posters will each Avenger have completed?

We are told in the question:

Number of days = x

Number of posters = y

Ms. Ironperson has completed 12 posters and will complete 6 more per day.

y = 12 + 6x.......... Equation 1

Mr. Thoro has not started yet but can make 12 per day.

y = 12x................Equation 2

Hence, the equations needed to solve this problem:

y = 12 + 6x.......... Equation 1

y = 12x .......... Equation 2

We substitute 12x for y in equation 1

y = 12 + 6x.......... Equation 1

12x = 12 + 6x

Collect like terms

12x - 6x = 12

6x = 12

x = 12/6

x = 2

Hence, since x = the number of days,

The number of days = 2 days

Substitute 2 for x in Equation 2

y = 12x .......... Equation 2

y = 12 × 2

y = 24 posters.

Since y = number of posters,

number of posters = 24 posters.

We were told in the question that, when Mr Thoro catches up with Ms Ironperson, that would both have completed the same number of posters.

Hence, both Avengers would have completed 12 posters each.

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Answer:

A. No real solution

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C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

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Answer:

11 terms.

Step-by-step explanation:

The exponent indicates how many time you will multiply 7 by itself.

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Answer:

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17 less than 12 times a certain number (m) IS 23 less than six times the same
REY [17]

Answer:

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Step-by-step explanation:

12 times the number is 12m

17 less than that is 12m -17.

23 less than 6 times the number is 6m-23.

These expressions  are said to be the same:

  12m -17 = 6m -23

  6m -17 = -23 . . . . . . . . subtract 6m from both sides

  6m = -6 . . . . . . . . . . add 17 to both sides

  m = -1 . . . . . . . . . divide both sides by 6

The value of m is -1.

3 0
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