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Brut [27]
3 years ago
6

Find an equation that passes through the points (-3,4) and (5,-4)

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0
Y=-1x+1 is the correct answer because you take the two points and go y2-y1/x2-x1, then one you figure that out you take y=mx+b and input your values (you can choose either point) so your work should look like: -4-4/5-3=-8/8=-1. -4=-1 (5)+b
-4= -5+b
+5. +5
1=b
y=-1x+1
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Please help and thank you
s2008m [1.1K]

Answer:

<em>Answer is</em><em>(</em><em>-1</em><em>,</em><em>-1</em><em>)</em>

Step-by-step explanation:

given \: points \: are \:( 1, - 6) \: and \: ( - 3,4) \\ midpoint \:  =  (\frac{x1 + x2}{2} , \frac{y1 + y2}{2} ) \\  =(  \frac{1 - 3}{2} , \frac{ - 6 + 4}{2} ) \\  (\frac{ - 2}{2} , \frac{ - 2}{2} ) \\ midpoint \:  =   ( - 1, - 1)

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em><em> </em><em> </em><em> </em><em> </em>

3 0
3 years ago
40% of students on a field trip love the museum. If there are 20 students on the field trip, how many love the museum??
MariettaO [177]
Since there are 20 students on the field trip. Then 20 is the 100% total. Thus, to find how many people love the museum, we can simply multiply the total number of people by the percentage of interest, namely 40%

20*0.4 = 8
So, 8 people love the museum.
5 0
3 years ago
Solve the formula for the red variable (a). *<br> 23. P= a + b + c<br> Your answer
Len [333]

Answer:

33

Step-by-step explanation:

7 0
4 years ago
Solve the system of equations given below
yaroslaw [1]

Answer:

idek

Step-by-step explanation:

5 0
3 years ago
For 0 ≤ ϴ &lt; 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va
Black_prince [1.1K]

Answer:

3 solutions:

\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is tan(\frac{\theta}{2}) so let's focus on that part of the equation first.

We know that:

tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}

therefore:

cos(\frac{\theta}{2})\neq0

so we need to find the angles that will make the cos function equal to zero. So we get:

cos(\frac{\theta}{2})=0

\frac{\theta}{2}=cos^{-1}(0)

\frac{\theta}{2}=\frac{\pi}{2}+\pi n

or

\theta=\pi+2\pi n

we can now start plugging values in for n:

\theta=\pi+2\pi (0)=\pi

if we plugged any value greater than 0, we would end up with an angle that is greater than 2\pi so,  that's the only angle we cannot include in our answer set, so:

\theta\neq \pi

having said this, we can now start solving the equation:

tan(\frac{\theta}{2})=sin(\theta)

we can start solving this equation by using the half angle formula, such a formula tells us the following:

tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}

so we can substitute it into our equation:

\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)

we can now multiply both sides of the equation by sin(\theta)

so we get:

1-cos(\theta)=sin^{2}(\theta)

we can use the pythagorean identity to rewrite sin^{2}(\theta) in terms of cos:

sin^{2}(\theta)=1-cos^{2}(\theta)

so we get:

1-cos(\theta)=1-cos^{2}(\theta)

we can subtract a 1 from both sides of the equation so we end up with:

-cos(\theta)=-cos^{2}(\theta)

and we can now add cos^{2}(\theta)

to both sides of the equation so we get:

cos^{2}(\theta)-cos(\theta)=0

and we can solve this equation by factoring. We can factor cos(\theta) to get:

cos(\theta)(cos(\theta)-1)=0

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

cos(\theta)=0

\theta=cos^{-1}(0)

\theta={\frac{\pi}{2}, \frac{3\pi}{2}}

Equation 2:

cos(\theta)-1=0

we add a 1 to both sides of the equation so we get:

cos(\theta)=1

\theta=cos^{-1}(1)

\theta=0

so we end up with three answers to this equation:

\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}

7 0
3 years ago
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