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3 years ago

The line that is the perpendicular bisector of the segment whose endpoints are R(-1, 6) and S(5, 5)

2 answers:

8 0

I think what you are looking for is to indicate an ecuation for this. So lets do the procedure:
First you need to find the slope of this line.
(5 - 6)/(5 + 1) = -1/6 
So, The slope of the perpendicular line is the opposite reciprocal, which is 
+6/1.Now we know the slope of the line but still have to find the mid point of the line. 
That is why we are going to use the mid point equation 
x = (-1 + 5)/2 = 2 
y = (6 + 5)/2 = 11/2 
(y - 11/2) = 6(x - 2) 
The last thing you do is rearrange the equation to get y = 6x - 6.5
Hope this is what you were looking for

KengaRu [80]3 years ago

3 0

Answer:

On odyssey ware it's 12x-2y=13

Step-by-step explanation:

yup

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Write an expression that represents n plus 7 multiplied by 3.

diamong [38]

The expression that represents n plus 7 multiplied by 3 is
(n + 7)(3)
The entire expression, (n + 7) is multiplied by 3 since it would be inconvenient to state the expression as given above if what is meant was
(n + 7x3). It would have simply been n plus 21.

7 0

3 years ago

Read 2 more answers

Plz help ASAP need problem quickly thx

dezoksy [38]

Answer:

Step-by-step explanation:

2(3) + (3) = 9

3(3) - 2(3) = 3

6 0

3 years ago

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Simplify 45÷2×8-12+42 Please show working

Allushta [10]

Answer:

210

Step-by-step explanation:

Lets use BEDMAS (Brackets, Exponents, Division, Multiplication, Addition, Subtraction)

45 ÷ 2 = 22.5

22.5 × 8 = 180

180 - 12 = 168

168 + 42 = 210

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Have a great summer :)

6 0

2 years ago

Answer is <, >, or =.

Pavlova-9 [17]

Answer:

≥

Step-by-step explanation:

6 0

3 years ago

find the value of a and b for which the system of equations has infinitely many solutions : 2x + 3y = 7 ; (a-b)x + (a+b)y = 3a +

hodyreva [135]

Answer:

a= 5 and b=1

Step-by-step explanation:

To solve, we will follow the steps below:

2x + 3y = 7

(a-b)x + (a+b)y = 3a + b - 2

We should note that since it has infinitely many solutions then,

$\frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} } = \frac{c_{1} }{c_{2} }$

Hence

2/a-b = 3/a+b = 7/3a +b-2

2/a-b = 3/a+b

cross-multiply

3(a-b) = 2( a+b)

open the bracket

3a - 3b = 2a + 2b

collect like term

3a - 2a = 2b + 3b

a = 5b -------------------------------------------(1)

Similarly

3/a+b = 7/3a +b-2

cross-multiply

7(a+b) = 3(3a+b -2)

7a + 7b = 9a + 3b -6

take all the variables to the left-hand side of the equation

7a - 9a+ 7b-3b = -6

-2a + 4b = -6 ---------------------------------(2)

but a = 5b

substitute a= 5b in equation (2) and solve for b

-2(5) + 4b = -6

-10 + 4b = -6

add 10 to both-side of the equation

-10 + 10+ 4b = -6+10

4b = 4

divide both-side of the equation by 4

b = 1

substitute b= 1 in equation (1)

a = 5b

a =5(1)

a=5

Therefore, a= 5 and b=1

8 0

3 years ago