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wel
3 years ago
6

Deniah Bostic

Mathematics
1 answer:
tatiyna3 years ago
3 0

Answer: The answer is \frac{8}{10}

Step-by-step explanation:

it is 8/10 because if you look at how many nickels there are it is 8 and there are ten dimes it wants a ratio but as a fraction the ratio would be 8:10 and as a fraction it would be 8/10

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2. Identify the solution to the system of equations graphed below.
12345 [234]
(-4, 1)

the solution is always where the two points meet!! :)
7 0
3 years ago
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Whats the ninth term of 25 20 15 10 5
Kaylis [27]

Given:

The sequence is 25, 20, 15, 10, 5.

To find:

The ninth term of the given sequence.

Solution:

We have,

25, 20, 15, 10, 5

It is an AP because the difference between two consecutive terms are same.

Here,

First term (a) = 25

Common difference (d) = 20-25

                                       = -5

The nth terms of an AP is

a_n=a+(n-1)d

Where, a is the first term and d is the common difference.

Putting a=25, n=9 and d=-5 to get the 9th term.

a_9=25+(9-1)(-5)

a_9=25+(8)(-5)

a_9=25-40

a_9=-15

Therefore, the ninth term of the given sequence is -15.

3 0
3 years ago
7 = x + 12<br><br><br><br> A. -19 B. -5 C. 4 D. 5
Elena L [17]

You have to get "X" by itself so you subtract 12, 7-12 is -5 so "X" = - 5


5 0
3 years ago
Read 2 more answers
The nth term of a sequence is 3n^2- 1
Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

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3 years ago
I need help with a lot of work due by tomorrow... someone please help!!
Dmitrij [34]

answer:what can i help you with

Step-by-step explanation:

6 0
4 years ago
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