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Alexeev081 [22]
3 years ago
11

Graph the line for y+1=−3/5(x−4) on the coordinate plane. Move Line Undo Redo Reset

Mathematics
1 answer:
dsp733 years ago
7 0

Answer:

The equation of  line is shown below.

Step-by-step explanation:

The equation of line is

y+1=\frac{-3}{5}(x-4)

It is the point slope form of a line, therefore the slope of the line is \frac{-3}{5} and the line passing through (4,-1).

Rewrite the above equation in slope intercept form.

y=\frac{-3}{5}x+\frac{12}{5}-1

y=\frac{-3}{5}x+\frac{12-5}{5}

y=\frac{-3}{5}x+\frac{7}{5}

The point slope form of a line is

y=mx+b

Where m is the slope and b is y-intercept.

Therefore the slope of the line is \frac{-3}{5} and the y-intercept of the line is \frac{7}{5}.

Slope is defined as

m=\frac{Rise}{Run}

Since the slope is negative, therefore the run of the line is considered on the left side. The line rise by 3 and run by 5, therefore we will add 7 in y and subtract 5 from x.

The y-intercept is (0,\frac{7}{5}).

(0-5,\frac{7}{5}+3)=(-5,\frac{22}{5})=(-5,4.4)

Therefore we have three points (4,-1), (-5,4.4) and (0,1.4). Plot the point on the coordinate plane and connect them be a straight line.

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Each variable indicates different weights. Which weight can you find? Find it.
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Answer:

We can only be certain that <em>a</em> weighs 12.

There are infinitely many possiblities for <em>b</em> and <em>c</em>.

Step-by-step explanation:

We have the equation:

a+b+c+a+c+b+a+c=12+a+a+b+b+c+c+c

Each variable indicates a weight.

We would like to determine the weights of each variable (if possible).

First, we can rearrange the equation to acquire:

(a+a+a)+(b+b)+(c+c+c)=12+(a+a)+(b+b)+(c+c+c)

We can combine like terms:

3a+2b+3c=12+2a+2b+3c

Notice that both sides have 2<em>b</em> and 3<em>c</em>. Therefore, it is possible for us to cancel them since each nullify the other side. So, we will subtract 2<em>b</em> and 3<em>c</em> from both sides. This yields:

3a=12+2a

Therefore, we can solve for <em>a</em>. Subtract 2<em>a</em> from both sides:

a=12

Hence, the weight of <em>a</em> is 12.

Using the newly acquired information, we can go back to our simplified equation:

3a+2b+3c=12+2a+2b+3c

Since <em>a</em> is 12:

3(12)+2b+3c=12+2(12)+2b+3c

Evaluate:

36+2b+3c=12+24+2b+3c

Simplify:

36+2b+3c=36+2b+3c

We can subtract 36 from both sides:

2b+3c=2b+3c

As you can see, this is a true statement.

Since this is a true statement, there are infinitely many possible values for <em>b</em> and <em>c</em>.

Therefore, the only weight we are <em>certain</em> of knowing is weight <em>a</em> weighing 12.

8 0
3 years ago
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