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2 years ago

Find an equation for the line that passes throughthe points (1,-1) and (-3,5)

Mathematics

2 answers:

Sladkaya [172]2 years ago

5 0

y = -1.5(x) + 0.5

points: (1,-1) and (-3,5)

slope:

⇒ (y2-y1)/(x2-x1)

⇒ (5-(-1))/(-3-1)

⇒ 6/-4

⇒ -1.5

equation:

⇒ y -(-1) = -1.5(x-1)

⇒ y + 1 = -1.5(x) + 1.5

⇒ y = -1.5(x) + 0.5

Tju [1.3M]2 years ago

4 0

U can pick either point to finish

y - 1 = 2(x-1)
y= 2x-2 +1
y = 2x-1

or

y-5 = 2(x-3)
y = 2x-6+5
y = 2x-1

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2 years ago

Read 2 more answers

Am I correct? ( top question)

EastWind [94]

Answer:

Step-by-step explanation:

<1 and <2 are not equal to each other

Let the angle directly above angle 2 (on the right side of the line) be angle 3

<1 and <3 are corresponding angle, which means they are equal

<2 and <3 are supplementary angles since the form a line

<2 + <3 = 180

We know <1 = <3

<1 + <3 = 180

We are given <1 = 2x+12 and <2 = 3x+18

2x+12 + 3x+18 = 180

Combine like terms

5x+30 = 180

Subtract 30 from each side

5x+30-30 = 180-30

5x= 150

Divide each side by 5

5x/5 = 150/5

x=30

<1 = 2x+12

Substitute 30 in for x to find angle 1

= 2*30 +12

=60+12

= 72

6 0

3 years ago

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EleoNora [17]

Answer:

Step-by-step explanation:

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3 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$