All categories

4 years ago

Circles M and K are congruent, arc QR is congruent to arc LN and arc OP is congruent to arc VW. Find x.

Mathematics

1 answer:

SCORPION-xisa [38]4 years ago

6 0

Given, arc QR is congruent to arc LN and arc OP is congruent to arc VW.

And the expressions for each arc in the diagram also given as:

Arc QR = 2x - y, arc LN = 11 , arc OP= 10 and arc VW=5x+y.

Hence, we will get the system of equations as following:

Arc QR = Arc LN

2x - y = 11 ...(1)

Arc OP = Arc VW

5x + y = 10 ...(2)

We need to find the value of x. So, we can add the equations to eliminate y so that we can solve the equations for x. Therefore,

2x+5x = 11 + 10

7x = 21

$\frac{7x}{7} =\frac{21}{7}$ Divide each sides by 7.

So, x= 3

You might be interested in

Determine the gradient and the y-intersect point of:

maks197457 [2]

Answer:

gradient:

3x+2y=5

3(1)+2dy/dx=0

3+2dy/dx=0

2dy/dx=-3

dy/dx=-3/2.

y-intersect:

3x+2y=5

2y=5-3x

y=5-3x/2

5 0

3 years ago

How many dots are in the 4th step?

Mkey [24]

Answer:

Step-by-step explanation:

if you look at the diagram the dot goes up 1 each. It also makes another row which makes 4 rows total. Since 3 contains 7 dots and 3 rows we would do 8 in 4 rows. 8 x 4 is 32.

Hope this helped!

7 0

2 years ago

Read 2 more answers

What is the area of this figure? Enter your answer in the box.

Basile [38]

Luckily for us, the diagram already divided this figure into separate polygons. What I will be explaining is basically the addition of the areas of all the separate polygons. The area of the uppermost triangle is:

1/2 x b x h
= 1/2 x 20 x 8
(the base is 20, because in a parallelogram, opposite sides are congruent)
=10 x 8
= 80 in. squared

The next polygon we will be taking the area of is the parallelogram with the base length of 20 and the height of 16.

Area = b x h
= 20 x 16
= 320 in. squared

Now all we have left to do is add the two areas to obtain the total area.

Total Area = 320 + 80 = 400 in. squared

3 0

3 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

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

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .