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3 years ago

What is the relationship of the line 3x - y = 5 to the line 6x - 2y = -7?

Mathematics

1 answer:

Marta_Voda [28]3 years ago

5 0

Answer:

Parallel

Step-by-step explanation:

in a Standard from equation, Ax+By=C

slope=-A/B

The slope of first line is -3/-1=3

The slope of second line is -6/-2=3

Same slopes but different C values mean they r parallel.

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Solve the equation -5x+31+3x=3

Rus_ich [418]

Answer:

Step-by-step explanation:

-5x+31+3x=3

move everthing to one side

-5+31+3x-3=0

add and subtract common terms

-2x+28=0

divide across by common denominator

-x+14=0

Solve for x

x=14

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3 years ago

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3 years ago

Line CD contains points A (4, 6) and B (−2, 6). The slope of line CD is (4 points)

lbvjy [14]

Answer:

Step-by-step explanation:

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m= (y2-y1)/(x2-x1)

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2 years ago

(11 – 9)' – 2 x 4=<br> Please help

PIT_PIT [208]

Answer: 2 − 8x

Step-by-step explanation:

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3 years ago

Read 2 more answers

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

1 year ago