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3 years ago

What is the equation of the line that passes through the points (1, –1) and (3, –7)

Mathematics

2 answers:

olganol [36]3 years ago

8 0

Answer:

-3x+2=y

Step-by-step explanation:

u need to find the slope intercept formula which mx+b=y

m=slope

b=y-intercept

first u should find the slope which is y2-y1/x2-x1

point 1:(1,-1)

point 2:(3,-7)

-7+1/3-1=-6/2=-3

m=-3

now replace the variables

-3(1)+b=-1

-3 times 1=-3

-3+b=-1

add 3 to both sides

b=2

now the equation is -3x+2=y

Akimi4 [234]3 years ago

6 0

Answer:

y = -3x + 2

Step-by-step explanation:

To find the slope of an equation you can do (y2-y1)/(x2-x1). Plug the values in from the equation and you get (-1+7)/(1-3), or 6/-2. This means the slope is -3. Then you plug the values of y and x into an equation y = -3x + b, and you get -1 = (-3)1 + b. Solve for b. -1 = -3 + b, (add 3 to each side) 2 = b. Then plug b back into your equation and you get y = -3x + 2

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Choose 3 values that would make this inequality true 28 + x > 42

tester [92]

Answer:

100, 15, 24 I think that is it

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Value of x for 12(x-2)+3x=1/2(x+6)+2

Damm [24]

Answer:

The value of x = 2

Step-by-step explanation:

Given the equation

$12\left(x-2\right)+3x=\frac{1}{2}\left(x+6\right)+2$

Expand 12(x-2) = 12x - 24

Expand 1/2(x+6) = 1/2x + 3

so the equation becomes

$12x-24+3x=\frac{1}{2}x+3+2$

simplifying

$15x-24=\frac{1}{2}x+5$

$15x=\frac{1}{2}x+29$

multiplying both sides by 2

$30x = x + 58$

subtracting x from both sides

$30x - x = 58 - x$

$29x = 58$

divide both sides by 29

$\frac{29x}{29}=\frac{58}{29}$

simplifying

$x = 2$

Therefore, the value of x = 2

4 0

3 years ago

This 1 seems really complicated

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;

4y − x + 4 = 0 ;
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"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________

y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :

→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
_________________________________________________

→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

a(b + c) = ab + ac ; AND:

a(b − c) = ab <span>− ac .
_________________________________________________
As such:

We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
______________________________________

→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ;   → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0)  − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________

7 0

3 years ago

How do I graph y = 1/5 cos x

MariettaO [177]

The answer is here. Hope I did help you.