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3 years ago

Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35

Mathematics

1 answer:

Shkiper50 [21]3 years ago

8 0

Step-by-step explanation:

<h2><em><u>concept :</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2><em><u>5y + 4x = 35</u></em></h2><h2 /><h2><em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>

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2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

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Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

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Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35 ​

Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35