If the figures/solids are similar, the ratio of the areas is equal to the square of the ratio of the corresponding sides. Let x be the area of the second figure. The equation that would best allow us to answer this item is,
(394.2 in² / x in²) = (10.9 / 21.8)²
The value of x from the equation is 1576.8 in².
Thus, the nearest answer is the second choice.
First make and equation. Since you want to find the number of the day where both charge the same total amount, you would make the equation:
let x= the number of days
110+46(x)=70+54(x)
Solve:
110-70+46(x)=70-70+54(x)
40+46(x)=54(x)
40+46(x)-46(x)=54(x)-46(x)
40=8(x)
40/8=8x/8
5=x
Therefore your answer would be 5 days
Answer:
-5, multiplicity 3; +9, multiplicity 2; -1
Step-by-step explanation:
The roots of f(x) are those values of x that make the factors be zero. For a factor of x-a, the root is x=a, because a-a=0. If the factor appears n times, then the root has multiplicity n.
f(x) = (x+5)^3(x-9)^2(x+1) has roots ...
- -5 with multiplicity 3
- +9 with multiplicity 2
- -1
Answer:
The answer is down below
Step-by-step explanation:
90
The calendar obviously has an integral number of years and months in 400 years. If it has an integral number of weeks, then it will repeat itself after that time. The rules of the calendar eliminate a leap year in 3 out of the four century years, so there are 97 leap years in 400 years. The number of excess days of the week in 400 years can be found by ...
(303·365) mod 7 + (97·366) mod 7 = (2·1 + 6·2) mod 7 = 14 mod 7 = 0
Thus, there are also an integral number of weeks in 400 years.
The first day of the week is the same at the start of every 400-year interval, so the calendar repeats every 400 years.
