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4 years ago

(PICTURE) Which equation is a point slope form equation for line AB ?

Mathematics

1 answer:

Gwar [14]4 years ago

4 0

The last one : y - 5 = $\frac{3}{4}$ (x - 6)

the equation of a line in point- slope form is

y - b = m(x - a )

where m is the slope and (a, b ) a point on the line

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = A(- 2, - 1 ) and (x₂, y₂ ) = (B(6, 5 )

m = $\frac{5+1}{6+2}$ = $\frac{6}{8}$ = $\frac{3}{4}$

using (a, b ) = (6, 5 ), then

y - 5 = $\frac{3}{4}$ (x - 6 ) ← in point- slope form

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Need help!!!!!!!!!!!!Please

Vikentia [17]

Plz help I’m getting timed

4 0

3 years ago

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

5 0

4 years ago

SYSTEM OF LINEAR EQUATIONS

Effectus [21]

Answer:

Step-by-step explanation:

x-y=11

2x+y=19

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

x=y+11

2(y+11)+y=19

2y+22+y=19

3y+22=19

3y=19-22

3y=-3

y=-3/3

y=-1

x-(-1)=11

x+1=11

x=11-1

x=10

Answer: x=10, y=-1. (10, -1).

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

-6x+5y=1

6x+4y=-10

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

9y=-9

y=-9/9

y=-1

6x+4(-1)=-10

6x-4=-10

6x=-10+4

6x=-6

x=-6/6

x=-1

Answer: x=-1, y=-1. (-1, -1).

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

2x+9y=25

-4x-9y=-23

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

-2x=2

x=2/-2

x=-1

2(-1)+9y=25

-2+9y=25

9y=25-(-2)

9y=25+2

9y=27

y=27/9

y=3

Answer: x=-1, y=3. (-1, 3).

4 0

3 years ago

Read 2 more answers

If you have the equation, y=2x+10 what’s x if y=24

jolli1 [7]

Answer: x=7

24=2x+10

24-10=14

14=2x

14/2=7

6 0

3 years ago

Please help with #12 im confused<br> that is the key but i dont get it

Svetllana [295]

Given Equation

$\boxed{ \tt \: (8 \sqrt{3} - 2 \sqrt{2} )(8 \sqrt{3} + 2 \sqrt{2})}$

Step by step expansion:

$\dashrightarrow \sf \: (8 \sqrt{3} - 2 \sqrt{2} )(8 \sqrt{3} + 2 \sqrt{2})$

$\\ \\$

$\dashrightarrow \sf \: (8 \sqrt{3})^{2} - (2 \sqrt{2} {)}^{2}$

Reason:

(a + b)(a - b) = a²- b²

$\\ \\$

$\dashrightarrow \sf \: (8 {}^{2} \times { \sqrt{3} }^{2} )- (2 \sqrt{2} {)}^{2}$

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$\dashrightarrow \sf \: (64 \times 3 ) - (2 \sqrt{2} {)}^{2}$

$\\ \\$

$\dashrightarrow \sf \: (64 \times 3 ) - (2 {}^{2} \times \sqrt{2} {}^{2} {)}$

$\\ \\$

$\dashrightarrow \sf \: 192- 8$

$\\ \\$

$\dashrightarrow \sf \: 184$

$\\ \\$

~BrainlyVIP ♡

4 0

3 years ago

Read 2 more answers