In circle P with MLNRQ =60, find the mLNPQ

(will give BRAINLIEST)The formula for the perimeter of a rectangle Is P = 2(1+w). Part A. Rewrite the formula for the perimeter

shepuryov [24]

A. W = P/2 - 1

B. First you need to divide both sides by 2 to get rid of the 2 on the side of the w. Then you need to take 1 away from both sides.

Whatever you do to one side, you must do to the other.
Hope this helps x

4 0

3 years ago

A student solved this problem and said the answer is 5 pounds. Todd has two packages of hamburger. One package weighs pound and

Gnesinka [82]

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7 0

3 years ago

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Write the equation of the perpendicular bisector of . Point B is located at (–3, 1), and point C is located at (1, 5).

maxonik [38]

The equation of perpendicular bisector of BC is

$y=-x+2$

Further explanation:

A perpendicular bisector passes through the mid-point of a line so to get an ordered pair from which the perpendicular bisector will pass we need to find the mid-point of BC.

$Mid-point = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} )\\Here\\(x_1,y_1) = (-3,1)\\(x_2,y_2) = (1,5)\\Putting\ values\ in\ formula\\M = (\frac{-3+1}{2} , \frac{1+5}{2} )\\= (\frac{-2}{2} ,\frac{6}{2})\\= (-1, 3)$

So the perpendicular bisector will pass through (-1,3)

Let m1 be the slope of BC

Then

$m_1 = \frac{y_2-y_1}{x_2-x_1} \\=\frac{5-1}{1+3} \\=\frac{4}{4}\\ =1$

Let m2 be the slope of perpendicular bisector

We know that the product of slopes of perpendicular lines is -1.

$m_1.m_2 = -1\\1*m_2 = -1\\m_2 = -1$

The standard form of point-slope form of line is:

$y=mx+b$

Putting the value of slope of bisector we get

$y=(-1)(x)+b\\y=-x+b$

To find the value of b, put the ordered pair (-1,3) in the equation

$3 = -(-1) +b\\3= 1+b\\3-1 =b\\b=2\\$

Putting the values of m and b in standard form

$y=-x+2$

Hence the equation of perpendicular bisector of BC is

$y=-x+2$

Keywords: Perpendicular bisector, Point-slope form

Learn more about point-slope form at:

#LearnwithBrainly

6 0

3 years ago

Pre-Calc help:

stepladder [879]

Answer:

$8x^2+8y^2-79x-32y+95=0$

Step-by-step explanation:

Let the equation be : $x^2+y^2+2gx+2fy+c=0$

The point (1,1) must satisfy this equation.

$1^2+1^2+2g*1+2f*1+c=0\\2g+2f+c=-2---(1)$

Similarly, the point (1,3) must satisfy:

$1^2+3^2+2g*1+2f*3+c=0\\1+9+2g+6f+c=0\\2g+6f+c=-10---(2)$

Also for the point (9,2), we have:

$9^2+2^2+2g*9+2f*2+c=0\\81+4+18g+4f+c=0\\18g+4f+c=-85---(3)$

Solving the equations (1), (2) and (3) simultaneously, we get:

$g=-\frac{-79}{16},f=-2,c=\frac{95}{8}$

We Substitute this values to get:

$x^2+y^2+2(\frac{-79}{16})x+2(-2)y+\frac{95}{8}=0$

$x^2+y^2-(\frac{79}{8})x-4y+\frac{95}{8}=0$

Multiply through by 8 to get:

$8x^2+8y^2-79x-32y+95=0$

8 0

3 years ago

A number is randomly selected from {15, 16, 17, 18, 19, 20}. A is represented by the event that the number picked is less than o

Tatiana [17]

1/3
the denonator is 6 because that is the number of numbers there are, and the numerator is 2 because there are 2 numbers less than or equal to 16. you can simplify

5 0

3 years ago

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