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2 years ago

A line has the equation 3x ? 4y = 1. Choose the equation of a line that is parallel to the given line.

Mathematics

2 answers:

cupoosta [38]2 years ago

7 0

Answer:

Find a line which also has 3/4 as the slope or 3x - 4y in standard form.

Step-by-step explanation:

If the line is 3x - 4y = 1 then the line which is parallel will have the same coefficients of x and y. Parallel lines never cross and to ensure this have the same slope. The slope is a ratio which can be solved for in an equation using the coefficients of x and y. Here the slope is:

3x - 4y = 1

-4y = -3x + 1

y = 3/4x - 1/4.

Find a line which also has 3/4 as the slope or 3x - 4y in standard form.

svet-max [94.6K]2 years ago

5 0

Answer: y= 3/4x +5

Step-by-step explanation: just did it

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Step-by-step explanation:

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3 years ago

What is the value of y in y-5=3-9(y+2)

Irina18 [472]

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3 years ago

If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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3 years ago