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3 years ago

Find the equation of the line that passes thru (-1,-5) and is parallel to 3x+y=-5. Show work.

Mathematics

1 answer:

sertanlavr [38]3 years ago

3 0

Answer:

$y=-3x-8$

Step-by-step explanation:

slope-intercept form:

$y=mx+b$

Start with the given equation, and to find the slope, convert it to slope-intercept form by solving for y.

$3x+y=-5$

In this case, all you need to do is subtract 3x from both sides.

$y=-3x-5$

Here, the slope is -3. The slope of a parallel line will also be -3.

Now, we have 3 known variables. We have the slope and the x and y of a known point. Use those 3 knowns to solve for the 1 unknown, the y-intercept.

$y=mx+b\\-5=(-3)(-1)+b\\-5=3+b\\-8=b\\b=-8$

Finally, with the slope and the y-intercept, you can write the equation of the line:

$y=-3x-8$

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so the answer is "the probability is 5/6 chance"

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Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

8 0

3 years ago

Solve for x:<br> -14x > -3

Alexus [3.1K]

Answer:

x>11

Step-by-step explanation:

-14x>-3

+14 +14

x > 11

6 0

3 years ago