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3 years ago

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A cable company charges $95 for installation plus $45 per month. If you paid $455 over a period of time, how many months did you

use that cable company?

1 answer:

alexira [117]3 years ago

5 0

Answer: 8 months

Step-by-step explanation:

m= number of months

45m+95=455

minus 95 over

45m=360

divide by 45

m=8

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A pet store holds training classes for dogs. The store has two different training programs. The first program charges a $35 memb

GrogVix [38]

The programs cost the same for 7 classes.

Given, no. of classes is represented by c, no. of dogs is represented by d, no. of hours is represented by h, the no. of programs is represented by p and the total cost is represented by t.

The first program charges $35 membership fee and $5 for each class.

The second program charges only $10 for each class.

Now the total cost say $t_{1}$ , for first program can be written in equation form as,

$t_{1} =35+5c$ , here c is the no. of classes.

Similarly, the total cost say $t_{2}$ , for second program can be written in equation form as,

$t_{2} =10c$ .

We have to find out the no. of classes for which the two programs cost the same.

according to the question,

$t_{1} =t_{2}$

$35+5c=10c\\10c-5c=35\\5c=35\\c=7$

Hence for 7 of classes the programs cost the same.

For more details follow the link:

brainly.com/question/4042361

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2 years ago

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Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

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2 years ago

Lacy has a $1000 bond with a 6.2% coupon that she purchased for $1025. What is the yield of this bond? Round your answer to 2 de

Ede4ka [16]

So the question is asking what is 6.2% of 1000? Then rounding our decimals?

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3 years ago

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2y = x + 3 5y = x - 7 What is the solution set of the given system?

prohojiy [21]

<span>2y = x + 3 then x = 2y - 3
5y = x - 7 then x = 5y + 7

so
</span>2y - 3 = 5y + 7
2y - 5y = 7 + 3
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y = -10/3
y = -3 1/3

<span>2y = x + 3
</span>2 (-10/3) = x + 3
-20/3 - 3 =x
-29/3 = x
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or
x = -9 2/3

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3 years ago

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10 more than 405. 4 hhundred 0 tens 5 ones

vova2212 [387]

(4 hundred + 0 tens + 5 ones) + (1 ten) = 4 hundred + 1 ten + 5 ones

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4 years ago