NO.,the given measures can not be the lengths of the sides of a triangle
Step-by-step explanation
The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
so, Find the range for the measure of the third side of a triangle given the measures of two sides.
here given measures are 2,2,6
2+2 = 4 which is less than the third side 6
= 4 < 6
This not at all a triangle.
Hence, the given measures can not be the lengths of the sides of a triangle
Answer:
Radius=9
Diameter=18
Step-by-step explanation:
c had the same ? then got it right
Answer:
the answer is 8.5
Step-by-step explanation:
divide 85 by 10 to get 8.5 for all the dimension
<h3>
<u>Answer:</u></h3>

<h3>
<u>Step-by-step explanation:</u></h3>
A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.
The inequality given to us is :-

Let's plot a graph to see its interval . Graph attached in attachment .
Now we can see that the Interval notation of would be ,
![\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Corange%20%5Ctt%20%5Cpurple%7B%5Cleadsto%7Dy%20%5Cin%20%5B-2%2C-1%5D%20%7D%7D)
<h3>
<u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>