Answer:
Step-by-step explanation:
he has 4.80 in dimes and quarters....dimes (d) are worth 0.10 and quarters (q) are worth 0.25
0.10d + 0.25q = 4.80
he has 6 more dimes then quarters
d = q + 6
====================
d = q + 6...so we sub q + 6 in for d, back into the other equation
0.10(q + 6) + 0.25q = 4.80
0.10q + 0.60 + 0.25q = 4.80
0.35q = 4.80 - 0.60
0.35q = 4.20
q = 4.20 / 0.35
q = 12 <====== there is 12 quarters
d = q + 6
d = 12 + 6
d = 18 <===== there is 18 dimes
6 gallons, if there is 216 gallons in 36 tubs you have to take 216÷36 to get 6 gallons. To check your answer you can take 36 × 6
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.
<h3>
Answer: x = 3</h3>
To get that answer, we replace y with 8 and then isolate x.
5x-7 = y
5x-7 = 8
5x = 8+7
5x = 15
x = 15/5
x = 3
This is so hard not gonna lie :0
