Answer:
c
Step-by-step explanation:
Two sum of any two side lengths of a triangle will always be greater than the length of the remaining side
a doesn't work because 8+12=20 and 20 is not greater than 20
b doesn't work because 4+5=9 and 9<10
d doesn't work because 6+6=12 and 12 is not greater than 12
c is the only option where you can take any two sides, add them, and the sum will be greater than the remaining side.
For example, for c,
6+8=14, and 14>10
8+10=18, and 18>6
10+6=16, and 16>8
Answer:
Step-by-step explanation:
Let's apply the formula (x+y)² = x² + 2xy + y²
Here, x = -a and y = b
So,
= (-a)² + 2(-a)(b) + (b)²
= a² - 2ab + b²
Hence, it has been proved that (-a + b)² = a² - 2ab + b².
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Answer: 6
Steps:
7 + 11 + 0 = 18
18/3 = 6
Answer:
Step-by-step explanation:
- sqrt(39) and square root 47
are the limits. The - square root of 39 is smaller than - 6. So the integer to use here is - 6
sqrt (47) = 6.686. Here the square root is larger than the closest integer.
The integer to use is 6
- 6 + 6 = 0
