The triangle pay $32 more for that day than it paid per day during the first period of time.
Step-by-step explanation:
The given is,
Triangle Construction pays Square Insurance $5,980
To insure a construction site for 92 days
To extend the insurance beyond the 92 days costs $97 per day
Triangle extends the insurance by 1 day
Step:1
Insurance per day from the 92 days period,

Where, Total insurance for 92 days = $ 5,980
Period = 92 days
From the values, equation becomes,

= $ 65 per day
Step:2
Insurance per day after the 92 days,
= $ 97
Amount Pay for that day than it paid per day during the first period of time,

= $32
Result:
The triangle pay $32 more for that day than it paid per day during the first period of time, if the Triangle Construction pays Square Insurance $5,980
to insure a construction site for 92 days and to extend the insurance beyond the 92 days costs $97 per day.
Answer:
24%
Step-by-step explanation:
I think this is right. im sorry if its not but i went to a percent calculator and checked my answer. hope this helped :)
Answer:
You firstly multiply 25,5cm with 20cm equals to 510cm then you divide it with 150 equals 3,4cm
Step-by-step explanation:
(25.5cm × 20cm) ÷ 150cm
≈510cm ÷150cm
≈3,4
Answer:
∴Given Δ ABC is not a right-angle triangle
a= AB = √45 = 3√5
b = BC = 12
c = AC = √45 = 3√5
Step-by-step explanation:
Given vertices are A(3,3) and B(6,9)

AB = 
Given vertices are B(6,9) and C( 6,-3)
= 
BC = 12
Given vertices are A(3,3) and C( 6,-3)

AC² = AB²+BC²
45 = 45+144
45 ≠ 189
∴Given Δ ABC is not a right angle triangle
It is $7. Anything average 5 you need to round up