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3 years ago

Find the equation of the line in slope intercept form that passes through (0,4) and is parallel to the line 4x-3y=12?

Mathematics

2 answers:

Lynna [10]3 years ago

6 0

Answer:

y=−4/3x+8

Step-by-step explanation:

y=mx+b

4x+3y=12→y=−4/3x+4

y=−4/3x+b

4)=−4/3(3)+b→b=8

y=−4/3x+8

Drupady [299]3 years ago

5 0

Answer:

y=−4/3x+8

Step-by-step explanation:

Let’s say that the parallel line takes on this form.

y=mx+b

If the line is parallel, that means that the m of the two lines are equal.

If we solve the line we were given for y, we find…

4x+3y=12→y=−4/3x+4

so, we know that our line that we’re trying to find must be of the form

y=−4/3x+b

Now we can plug in a point to fix that b part.

(4)=−43(3)+b→b=8

So, that means…

y=−4/3x+8

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3 years ago

when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.

sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = $a^2$

6 is subtracted from the square of a number = $a^2 - 6$

5 times the number = $5 \times a$

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

$a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0$

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in this problem,

$a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6$

Substituting the values in above quadratic formula, we get

$\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}$

We have two solutions for "a"

$\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}$

We have asked negative solution. So a = -1

Thus the negative solution is -1

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