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3 years ago

Which is the equation of a line parallel to the line with equation y=3x-4

Mathematics

1 answer:

prisoha [69]3 years ago

7 0

Answer

y=−3x+10

The given equation is in slope-y intercept form. The slope is the coefficient of x…which is -3, and the y-intercept is the constant term…+4.

Parallel lines have the same slope. So the correct equation should have a slope of -3. The only change will be the constant term (y-intercept) which needs to be +10. which gives us the equation….y =-3x+10.

Step-by-step explanation:

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2 years ago

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vekshin1

Hello,

The correct answer is D) 2

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3 years ago

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3 years ago

Read 2 more answers

Find sin 2a and cot 2a:

geniusboy [140]

Answer:

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cot(2\alpha ) = \frac{16511}{18720}$

Step-by-step explanation:

$\text{if } cos(\alpha)=\frac{12}{13}\\\text{That must mean we have a triangle with base 12, and hypotenuse 13.}\\\text{Using Pythagoras, we can determine the base of the triangle must be 5.}\\a^2+b^2=c^2 \text{, where c is the hypotenuse and a, b are the two other sides.}\\c^2-b^2=a^2\\\sqrt{c^2-b^2}=a\\\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\\\text{Therefore, }sin(\alpha) = \frac{5}{12}\\sin(2\alpha)=2sin(\alpha )cos(\alpha)\\\text{(From double angle formulae identities)}\\$

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cos(2\alpha )=cos^2(\alpha)-sin^2(\alpha)\\cos(2\alpha )=(\frac{12}{13})^2-(\frac{5}{12})^2=\frac{16511}{24336}\\cot(2\alpha)=\frac{cos(2\alpha)}{sin(2\alpha)}=\frac{\frac{16511}{24336}}{\frac{10}{13}}=\frac{16511}{18720}$

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3 years ago

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steposvetlana [31]

Answer:

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Step-by-step explanation:

sowwy

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2 years ago