m^2 -11m -60 = 0
m^2 -11m =60
(m^2 -11m + 121/4) = 361/4
(m - 11/2)^2 = 361/4
m - 11/2 = + or - 19/2
m = 11/2 + 19/2, 11/2 - 19/2
m = 15, -4
---------------------------------------------------
m^2 -11m -60
(11 +- Square root (121 +240))/2
11/2 +- 19/2
m= 15, -4
----------------------------------------------------
m^2 -11m -60
(m - 15) (m + 4)
m = 15, -4
10
13.50 per set
+.45 per day value
she got 20 sets
5 days after would be
13.50 ×20=270
value of 20 sets at 13.50 a set on day one was 270.00
.45×20=9.00
the 20 sets gain 9 dollars a day
9×5=45
9 dollars a day times 5 days is 45 dollars
45+270=315
11
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.
Answer:
Step-by-step explanation:
V=Tr^2h
V / T = r^2 h
V / Th = r^2
√V / Th
6 = x + 2
Subtract 2 from both sides
6 - 2 = x + 2 - 2
4 = x
