Answer:
The answer is "MS and QS".
Step-by-step explanation:
Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.
As NR and MQ bisect each other at S
⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ
In ΔMNS and ΔQNS
MN=QN (∵ MNQ is isosceles triangle)
∠NMS=∠NQS (∵ MNQ is isosceles triangle)
MS=SQ (Given)
By SAS rule, ΔMNS ≅ ΔQNS.
Hence, segments MS and SQ are therefore congruent by the definition of bisector.
The correct option is MS and QS
Following are the solution parts for the given question:
For question A:
In the given question, we calculate 
of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:
Using the t table we calculate 
When of the confidence interval:
So confidence interval for the mean weight of shipped homemade candies is between 
.
For question B:
Here we need to calculate confidence interval for the true proportion of all college students who own a car which can be calculated as
Using the Z-table we found 
therefore the confidence interval for the genuine proportion of college students who possess a car is
So the confidence interval for the genuine proportion of college students who possess a car is between 
For question C:
- In question A, We are certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
-
In question B, We are positive that the true percentage of college students who possess a car is between 0.28 and 0.34.
Learn more about confidence intervals:
brainly.in/question/16329412
Check the picture below.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
![\bf -------------------------------\\\\ d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2} \\\\\\ d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\ -------------------------------\\\\ d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\ -------------------------------\\\\ d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2} \\\\\\ d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%284-2%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%282%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B13%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%283-2%29%5E2%2B%28-2-4%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B37%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-2-3%29%5E2%2B%5B-3-%28-2%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-5%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
![\\\\ -------------------------------\\\\ d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2} \\\\\\ d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-1-%28-2%29%5D%5E2%2B%5B2-%28-3%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B2%29%5E2%2B%282%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%281%29%5E2%2B%285%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
so, those are their lengths, sum them all up, that's the polygon's perimeter.
Answer:
1. A(1;5), B(10;23), slope of 2
2. A(10;9), B(4;0), slope of 1.5
3. A(-35/4;6), B(9;-35/6), slope of -2/3
4. A(5;18), B(25,22), slope of 0.2
5. A(2/9;1/3), B(2/3,-1/3), slope of -1.5
Step-by-step explanation:
Substitute values with correct variable and slopemis coefficient of x when x and y-int. are on 1 side, y is on other side and has coefficient of 1. Hope it works!
That would be A...they have to have a constant ratio
