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3 years ago

Write an equation in standard form that has a slope of -1 and passes through the point (-5, -1).

Mathematics

2 answers:

joja [24]3 years ago

8 0

Answer:

y = -x-6

Step-by-step explanation:

For this problem we use point-slope formula:

y-y1 = m (x2-x1)

When we plug in our values, it looks like this:

y-(-1) = -1 (x-(-5))

When we rearrange into slope-intercept form, we get this:

y = -x-6

HTH :)

Grace [21]3 years ago

3 0

Answer:

x + y = -6

Step-by-step explanation:

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Answer:

Step by Step Explanation:

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3 years ago

How do you convert 5y-x+4=0 into y=mx+b form?

Mashutka [201]

$5y-x+4=0$

~Add x to both sides

$5y+4=x$

~Subtract 4 from both sides

$5y=x-4$

~Divide both sides by 5

$\frac{5}{5}y= \frac{1x}{5} - \frac{4}{5}$

~Simplify

$y= \frac{1}{5} x- \frac{4}{5}$

~M= $\frac{1}{5}$

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3 years ago

What equation has a constant of proportionality equal to 4

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3 years ago

What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?

Ahat [919]

Answer:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form $x-c$ .

Since we are dividing by $x+2=x-(-2)$ , this is our setup for the synthetic division:

-2 | 1 7 9 -8

| -2 -10 2

______________

1 5 -1 -6

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

So $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

We can also do long division.

x^2+5x-1

____________________

x+2| x^3+7x^2+9x-8

-(x^3+2x^2)

-------------------

5x^2+9x-8

-( 5x^2+10x)

--------------------

-x-8

-(-x-2)

--------------

-6

So we see here we get the same quotient, $x^2+5x-1$ . and the same remainder, $-6$ .

Now let's check our result that:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

So I'm going to rewrite the right hand side as a single fraction:

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}$ .

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

Now let's focus on multiplying $(x+2)(x^2+5x-1)$ .

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

$x(x^2)=x^3$

$x(5x)=5x^2$

$x(-1)=-x$

$2(x^2)=2x^2$

$2(5x)=10x$

$2(-1)=-2$

---------------------------Combine like terms:

$x^3+(5x^2+2x^2)+(-x+10x)+-2$

$x^3+7x^2+9x-2$

So let's go back where we were in our check of $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ :

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}$

We have the exact same thing on both sides so we did good.

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