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3 years ago

Write the equation of a line that is parallel to y = -x + 7

Mathematics

1 answer:

Gnoma [55]3 years ago

4 0

Answer:

y= -x-3

Step-by-step explanation:

A line is parallel to another line if their slope is the same. In this case the slope is -1 (-x is basically -1x). So far we have y=-x+b (We dont know what b is yet). next we can insert the coordinate back into the equation y=-x+b where 1 is the y and -4 is the x.

We get...

y= -x+b

1= -(-4)+b

1=4+b

b=-3

Our final equation is y=-x-3

Hope this helped

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Answer:

9.9%

Step-by-step explanation:

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3 years ago

A line passes through the points (-6, 4) and (-2, 2). Which is the equation of the line?

cupoosta [38]

Answer:

$\displaystyle y=-\frac{1}{2}x+1$

Step-by-step explanation:

The equation of any line in slope-intercept form is:

y=mx+b

Being m the slope and b the y-intercept.

Assume we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Two points are given: (-6,4) and (-2,2). Calculating the slope:

$\displaystyle m=\frac{2-4}{-2+6}=\frac{-2}{4}=-\frac{1}{2}$

The equation of the line is, so far:

$\displaystyle y=-\frac{1}{2}x+b$

To calculate the value of b, we use any of the given points, for example (-6,4):

$\displaystyle 4=-\frac{1}{2}(-6)+b$

$\displaystyle 4=3+b$

Solving:

b = 1

The equation of the line is:

$\boxed{\displaystyle y=-\frac{1}{2}x+1}$

We can see none of the choices is correct.

8 0

3 years ago

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

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3 years ago