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3 years ago

Please help I’ll give brainliest

Mathematics

1 answer:

Hatshy [7]3 years ago

4 0

Help with what tho you need help with what I don’t see it

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What is the GCF of -2xy and -9y

NNADVOKAT [17]

Answer:

$- y(2x + 9)$

Step-by-step explanation:

The GCF doesn't have to be a number. In this case, it is -y because both the -2xy and -9y share it. It's also easier when you take out a negative. Once you take the GCF out, your new expression should be:

$- y(2x + 9)$

5 0

3 years ago

Help? Giving brainly!!

My name is Ann [436]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Please help quickly

labwork [276]

I believe it is twenty eight. Ten is two thirds of fifteen. Twenty eight is two thirds of forty two. But I'm just assuming tbh

6 0

2 years ago

If dB=4 and dc=6 find ad

Neporo4naja [7]

See the attached figure
DB = 4 and DC = 6 , We need to find AD

Using Euclid's theorem for the right triangle

∴ DB² = AD * DC

∴ 4² = AD * 6

∴ 6 AD = 16


∴ AD = 16/6 = 8/3 ≈ 2.67

7 0

2 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.

5 0

2 years ago