The longest possible integer length of the third side of the triangle is 6 < x < 28
The sum of any two sides must be greater than the third side for a triangle to exist
let the third side be x
x + 11 > 17 and x + 17 > 11 and 11 + 17 > x
x > 6 and x > - 6 and x < 28
The longest possible integer length of the third side of the triangle is 6 < x < 28
The length of the 3 sides of a triangle needs to always be among (however no longer the same) the sum and the difference of the opposite two sides. As an example, take the instance of two, 6, and seven. and. consequently, the third side period should be extra than 4 and less than 8.
Learn more about triangles here: brainly.com/question/1675117
#SPJ4
Answer:
Total number of arrangements = 1,663,200
Step-by-step explanation:
Given:
11 - letter
FIRECRACKER
F = 1
I = 1
R = 3
E = 2
C = 2
A =1
K = 1
Find:
Total number of arrangements = ?
Computation:
Note: Repeated letter will be avoid.
Total number of arrangements = 1,663,200
Answer:
i think its f :) double check tho
Step-by-step explanation:
First, distribute 2/3 to the values in the parentheses. This gives you y-1=2/3x+6.
To get y alone, we add 1 to both sides. This gets you y=2/3x+7.
y=mx+b
y=2/3x+7
It matches the standard form, so your final answer would be y=2/3x+7.
If you choose a blue marble first its the probability of 3/12 simplified to 1/3 and then the probability of choosing a red next is 5/11 as you didnt replace the one blue marble
