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viva [34]
2 years ago
8

Find an equation of the straight line that is perpendicular to the straight line x+2y=5 and that passes through the point (3,7)

Mathematics
1 answer:
Digiron [165]2 years ago
7 0

Answer: y = 2x + 1

Step-by-step explanation:  

Rewrite the equation in standard form:

x+2y=5

2y = -x + 5

y = -(1/2)x + (5/2)

A line perpendicular to this would have a slope that is the negative inverse of the reference line slope, -(1/2), here.  The new slope would be 2.  The y-intercept, (5/2), will change, so we'll just use b for the new y-intercept, and calculate it later:

y = 2x +b

We know that point (3,7) lies on the new line (it is a solution to the equation), so we can use this point to find the new y-intercept, b.

7 = 2*3 + b

b = 1

The perpendicular line is y = 2x + 1, and goes through point (3,7)

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Determine the zeros of f(x)=x^3+7x^2+10x-6
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x^3+7x^2+10x-6=x^3+3x^2+4x^2+12x-2x-6
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1,2,5 No 3,4 Yes

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What are the zeos of f(x)=(x-1)(x+6)
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The zeroes of the given polynomial is 1 and -6.

<h3>What is Polynomials?</h3>

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Here, given equation;

          f(x) = (x - 1)(x + 6)

for finding the zeros of the equation we need to put.....f(x) = 0

           0 = (x - 1)(x + 6)

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Thus, The zeroes of the given polynomial is 1 and -6.

Learn more about Polynomials from:

brainly.com/question/17822016

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