<h3>
Answer is -28</h3>
==================================================
Work Shown:
T(2) = 20 means the second term is 20
T(1) = 26 because we go backwards from what the rule says (subtract 6) to step back one term. Going forward, 26-6 = 20.
Since a = 26 is the first term and d = -6 is the common difference, the nth term is
T(n) = a + d*(n-1)
T(n) = 26 + (-6)(n-1)
T(n) = 26 - 6n + 6
T(n) = -6n + 32
Plugging n = 1 into the equation above leads to T(1) = 26. Using n = 2 leads to T(2) = 20.
Plug in n = 10 to find the tenth term
T(n) = -6n + 32
T(10) = -6(10) + 32
T(10) = -60+32
T(10) = -28
Answer:
bro just did this on a TGA
Step-by-step explanation:
A. no solutions and a contradiction
B. infinate solutions and an identity
C. one solution and niether
That would cost her $25.02
Unsure what the T/F statements are they are not included
Stokked 120
how many left?
300-120=180
180/300=x%
percent means parts out of 100
180/300=60/100=60%
60% left
Answer:
He has 11 quarters
Step-by-step explanation:
* Lets study the information in the problem to solve it
- The value of dimes and quarters is $6.35
- There are dimes and quarters
- The dime = 10 cents
- The quarter = 25 cents
* We must change the money from dollars to cents
∵ $1 = 100 cents
∴ $6.35 = 6.35 × 100 = 635 cents
- The number of dimes = 3 + 3 × number of quarters
* Let number of dimes is D and number of quarter is Q
∴ D = 3 + 3Q
∴ 10D + 25Q = 635
* Substitute the value of D from first equation in the second equation
∴ 10(3 + 3Q) + 25Q = 635 ⇒ open the bracket
∴ 10(3) + 10(3Q) + 25Q = 635
∴ 30 + 30Q + 25Q = 635 ⇒ collect like terms
∴ 30 + 55Q = 635 ⇒ subtract 30 from both sides
∴ 55Q = 605 ⇒ divide both sides by 55
∴ Q = 11
* He has 11 quarters
