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3 years ago

School started at 8:05 a.m and ended at 2:40p.m. the work below shows Erica's calculation of the length of the school day

Mathematics

1 answer:

harina [27]3 years ago

7 0

Answer:

B. She subtracted the end time from the start time and

D. She did not account for a.m. and p.m. in the start and end times.

Step-by-step explanation:

The school started at 8:05 a.m. and ended at 2:40 p.m.

And the length of the school day is calculated by Erica and it is shown in the attached photo.

See the attached photo.

Erica has done two errors which are

B. She subtracted the end time from the start time and

D. She did not account for a.m. and p.m. in the start and end times. (Answer)

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Solve the system of equations using the substitution method.

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y = -1 + 3/8x

2x - 5y = 6

Substitute the first equation into the second equation, since y is already by itself.

2x - 5(-1 + 3/8x) = 6

2x + 5 - 15/8x = 6

2x - 15/8x = 1

16/8x - 15/8x = 1

1/8x = 1 Multiply 8 on both sides to get x by itself

x = 8

Plug x into either of the equations.

y = -1 + 3/8(8)

y = -1 + 3

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2(8) - 5y = 6

16 - 5y = 6

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Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an

slega [8]

Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

x is the number of investors that contributed 4,000.

y is the number of investors that contributed 8,000.

Building the system:

There are 60 investors. So:

$x + y = 60$

In all, the partnership raised $348,000. So:

$4000x + 8000y = 348000$

I am going to simplify by 4000. So:

$x + 2y = 87$

Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

$1)x + y = 60$

$2)x + 2y = 87$

I am going to multiply 1) by -1. So we have

$1)-x - y = -60$

$2)x + 2y = 87$

By addition, the x are going to cancel each other

$-x + x - y + 2y = -60 + 87$

$y = 27$

For x:

$x + y = 60$

$x = 60-y = 60-27 = 33$

There were 33 $4,000 investors and 27 $8,000 investors.

2) Solve the system by row-reducing the corresponding augmented matrix.

$2x + y = 17$

$x + y = 13$

This system has the following augmented matrix:

$\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]$

To help the row reducing, i am going to swap the first with the second line:

$L1 L2$

So we have:

$\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]$

Now, reducing the first column.

$L2 = L2 - 2L1$

So we have:

$\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]$

Now we do:

$L2 = -L2$

And the matrix is:

$\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]$

Now to reduce the second column, we do:

$L1 = L1 - L2$

$\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right]$ .

So the solution is:

x = 4, y = 9.

3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?

I am going to say that x is the number of nickels and y is the number of dimes.

Each nickel is worth 5 cents and each dime is worth 10 cents.

Building the system:

There are 80 coins in all:

$x + y = 80$

They are worth $6.80. So:

$0.05x + 0.10y = 6.80$

Solving the system:

$1)x + y = 80$

$2)0.05x + 0.10y = 6.80$

I am going to divide 1) by -10, so we can cancel y. So:

$1)-0.10x - 0.10y = -8$

$2)0.05x + 0.10y = 6.80$

Adding:

$-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80$

$-0.05x = -1.2$ *(-100)

$5x = 120$

$x = \frac{120}{5}$

$x = 24$

Also

$x + y = 80$

$y = 80-x = 80-24 = 56$

There were 24 nickels and 56 dimes.

8 0

3 years ago