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)
Next step is to add the two equation to eliminate y so that we can solve the equations for x. Therefore,
2x+5x = 11 + 10
7x = 21
Divide each sides by 7.
So, x= 3
Now plug in x=3 in equation (2) to get the value of y.
5(3) + y = 10
15 + y =10
15 + y - 15 = 10 - 15 Subtracting 15 from each sides.
y = -5
So, x=3 and y =-5
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The equation of the graph in standard form is 2x + 3y = -6
<h3>How to determine the equation?</h3>
The attached figure represents the missing piece in the question
The graph passes through the points
(-3,0) and (0,-2)
The linear equation is then calculated using:

So, we have:

Evaluate the expression

Multiply through by 3
3y = -2x - 6
Add 2x to both sides
2x + 3y = -6
Hence, the equation in standard form is 2x + 3y = -6
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Answer:
D
Step-by-step explanation:
Let's review the distributive property.
It is
a(b+c) = ab + ac
Whenever we have a sum/difference multiplied by a number upfront, we need to eliminate the parenthesis. We do it by multiplying the constant with the terms of the sum/difference inside the parenthesis, each respectively.
Looking at 2 steps, we see Rob:
- eliminated the parenthesis by multiplying -5 with both "3x" and "+7"
- used the distributive property
From the answer choices, D is right.
Answer:
We find the length of each subinterval dividing the distance between the endpoints of the interval by the quantity of subintervals that we want.
Then
Δx= 
Now, each
is found by adding Δx iteratively from the left end of the interval.
So

Each subinterval is
![s_1=[-2,-3/2]\\s_2=[-3/2,-1]\\s_3=[-1,-1/2]\\s_4=[-1/2,0]](https://tex.z-dn.net/?f=s_1%3D%5B-2%2C-3%2F2%5D%5C%5Cs_2%3D%5B-3%2F2%2C-1%5D%5C%5Cs_3%3D%5B-1%2C-1%2F2%5D%5C%5Cs_4%3D%5B-1%2F2%2C0%5D)
The midpoints of the subintervals are

The points used for the
1. left Riemann sums are the left endpoints of the subintervals, that is

2. right Riemann sums are the right endpoints of the subinterval,

3. midpoint Riemann sums are the midpoints of each subinterval
