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monitta
2 years ago
10

What’s greater -15 1/2 or -20 3/4

Mathematics
1 answer:
andrezito [222]2 years ago
6 0
The answer is -15 1/2
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10 increased by twice Chrissy's height Use the variable c to represent Chrissy's height.
ahrayia [7]

Answer:

2C + 10

Step-by-step explanation:

3 0
3 years ago
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Identify a possible first step using the elimination method to solve the system and then find the solution to the system. 3x - 5
Nastasia [14]

Answer:

see explanation

Step-by-step explanation:

Given the 2 equations

3x - 5y = - 2 → (1)

2x + y = 3 → (2)

Multiply (2) by 5 will eliminate y when added to (1), that is

10x + 5y = 15 → (3)

Add (1) and (3) term by term

(3x + 10x) + (- 5y + 5y) = (- 2 + 15)

13x = 13 ( divide both sides by 13 )

x = 1

Substitute x = 1 into (2) for corresponding value of y

2 + y = 3 ⇒ y = 3 - 2 = 1

Solution is (1, 1)

7 0
3 years ago
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Joseph's lunch at a restaurant costs 13.00, without tax. He leaves the waiter a tip of 17% of the cost of the lunch, without tax
Musya8 [376]

Answer:

The cost before tax with tip is $15.21.

Step-by-step explanation:

To calculate percentage, take the percent (17%) and take its decimal form. This means 17% is 0.17 (like 50% would be 0.5). multiply the number you want the percentage of by the decimal form. 13*0.17=2.21. Add this to the cost before tax and before tip. 13.00+2.21=15.21. The cost before tax with tip is $15.21.

3 0
2 years ago
What is the value of n in the equation shown<br> below?<br> 2² x 2 = (24)
ANEK [815]

The answer I get


8=24
No solution
3 0
3 years ago
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}

6 0
2 years ago
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