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grandymaker [24]
3 years ago
12

What is the y-intercept of the line?

Mathematics
2 answers:
Luda [366]3 years ago
6 0

Answer:

1

Step-by-step explanation:

Blizzard [7]3 years ago
5 0
It’s either 1 or 1/2, I believe it’s 1 more though so try that :)
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△ABC = △DBC, AB= 10, AC=7 DC= ?
Furkat [3]

Answer:DC=7

Step-by-step explanation:

As it is written △ABC = △DBC.   So the corresonding legs (the legs which are equal ) are as follows:

AC=DC       AB=DB      BC= BC

So DC=7  the same like AC

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Which set of ordered pairs is made up of points on the graph of the function below? y=-3x+6
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3 years ago
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
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