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3 years ago

Which is the slope of the like represented by the equation 2x+3y=-12

2 answers:

6 0

$\text{The slope-intercept form:}\ y=mx+b\\m-slope,\ b-y-intercept\\\\2x+3y=-12\qquad|\text{subtract 2x from both sides}\\\\3y=-2x-12\qquad|\text{divide both sides by 3}\\\\y=-\dfrac{2}{3}x-4\\\\Answer:\ \boxed{m=-\dfrac{2}{3}}$

Burka [1]3 years ago

4 0

slope = - $\frac{2}{3}$

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 2x + 3y = - 12 into this form

subtract 2x from both sides

3y = - 2x - 12 ( divide all terms by 3 )

y = - $\frac{2}{3}$ x - 4 ← in slope- intercept form

with slope m = - $\frac{2}{3}$

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What are the ordered pairs of the

OlgaM077 [116]

Answer:

The two points solutions to the system of equations are: (2, 3) and (-1,6)

Step-by-step explanation:

These system of equations consists of a parabola and a line. We need to find the points at which they intersect:

$x^2-2x+3=-x+5\\x^2-2x+x+3-5=0\\x^2-x-2=0\\(x-2)(x+1)=0$

Since we were able to factor out the quadratic expression, we can say that the x-values solution of the system are:

x = 2 and x = -1

Now, the associated y values we can get using either of the original equations for the system. We pick to use the linear equation for example:

when x = 2 then $y=-(2)+5=3$

when x= -1 then $y=-(-1)+5=6$

Then the two points solutions to the system of equations are: (2, 3) and (-1,6)

6 0

3 years ago

I just need help with this problem

Oksanka [162]

+1/31 and -1/31
Hope this helps.

7 0

3 years ago

PLEASE HELP ME FIGURE THIS OUT 1......

MA_775_DIABLO [31]

Make sure your calculator is in "Deg" <em>(degrees)</em>.
Input "32"
Click on the "sin" button.

Answer: 0.5514

*********************************************************************

Answer: tan θ

<u>Step-by-step explanation:</u>

cotθ * sin²θ * sec²θ

= $\frac{cos}{sin} * \frac{sin*sin}{1} * \frac{1}{cos*cos} = \frac{sin}{cos}$ = tan

8 0

3 years ago

A screening test for a certain disease is used in a large population of people of whom 1 in 1000 actually have the disease. Supp

sertanlavr [38]

Answer:

$P(D/T)=5.05*10^{-6}$

Step-by-step explanation:

Let's call D the event that a person has the disease, D' the event that a person doesn't have the disease and T the event that the person tests negative for the disease.

So, the probability P(D/T) that a randomly chosen person who tests negative for the disease actually has the disease is calculated as:

P(D/T) = P(D∩T)/P(T)

Where P(T) = P(D∩T) + P(D'∩T)

So, the probability P(D∩T) that a person has the disease and the person tests negative for the disease is equal to:

P(D∩T) = (1/1000)*(0.005) = 0.000005

Because 1/1000 is the probability that the person has the disease and 0.005 is the probability that the person tests negative given that the person has the disease.

At the same way, the probability P(D'∩T) that a person doesn't have the disease and the person tests negative for the disease is equal to:

P(D'∩T) = (999/1000)*(0.99) = 0.98901

Finally, P(T) and P(D/T) are equal to:

P(T) = 0.000005 + 0.98901 = 0.989015

$P(D/T) = 0.000005/0.989015 = 5.05*10^{-6}$