I = / r where I = current and r = resistance
80 = k / 50 so
k = 400
so we have I = 400/r
when r = 40
I = 400/40 = 10 amps
A= p (1-r)^t
A material after 11 years
P current amount 260
R rate of decreases 0.0092
T time 11 years
A=260×(1−0.0092)^(11)
A=234.87 round
A=235
1 + 2^1 = 3
3 + 2^2 = 7
7 + 2^3 = 15
15 + 2^4 = 31
31 + 2^5 = 63
63 + 2^6 = 127
so the next two terms in the sequence 1, 3, 7, 15, 31, is 63 and 127.
Answer:
5
Step-by-step explanation:
sum the parts of the ratio, 3 + 4 = 7 parts
Divide the number of students by 7 to find the value of one part of the ratio
35 ÷ 7 = 5 ← value of 1 part of the ratio, then
3 parts 3 × 5 = 15 ← number of boys
4 parts = 4 × 5 = 20 ← number of girls
20 - 15 = 5
There are 5 more girls than boys
