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BlackZzzverrR [31]
3 years ago
15

Two lines that are perpendicular intersect to make 90 degree angles that are classified

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
8 0
Wouldn’t it be a right angle, I’m pretty sure it would be a right angle. Only perpendicular lines can intersect and create one.
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Step-by-step explanation:

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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
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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.

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