Shortest side (a) = 58
middle side (b) = 64
longest side (c) = 77
the 3 sides a + b + c = 199
b = a + 6
c = a + 19
substitute your new values for b & c into your original formula, so:
a + (a+6) + (a+19) = 199
3a + 25 = 199
3a = 174
a = 58
then substitute 58 into your b & c formulas to figure out the rest
b = a + 6 = 58 + 6 = 64
c = a + 19 = 58 + 19 = 77
Answer: The number is 26.
Step-by-step explanation:
We know that:
The nth term of a sequence is 3n²-1
The nth term of a different sequence is 30–n²
We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.
we get:
3n²- 1 = 30–m²
Notice that as n increases, the terms of the first sequence also increase.
And as n increases, the terms of the second sequence decrease.
One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.
if m = 1, then:
3n²- 1 = 30–1²
3n²- 1 = 29
3n² = 30
n² = 30/3 = 10
n² = 10
There is no integer n such that n² = 10
now let's try with m = 2, then:
3n²- 1 = 30–2² = 30 - 4
3n²- 1 = 26
3n² = 26 + 1 = 27
n² = 27/3 = 9
n² = 9
n = √9 = 3
So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.
the number is:
3*(3)² - 1 = 26
30 - 2² = 26
The number that belongs to both sequences is 26.
its y-axis for the first answer and the other one its, y=-x
(35 x 3) + (55 x 2)
105 + 102
answer = 207 minutes
Answer:
8 times 3 times 2+2 times 6 times 2+8 times 6 times 2
Step-by-step explanation:
