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Alborosie
3 years ago
15

Alguien me ayuda????

Mathematics
1 answer:
meriva3 years ago
4 0

What do you need help with?

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2(x-+2)+4=6x-4(-1+x)
sertanlavr [38]

2x-4+4=6x+4-4x

2x=2x+4

c there is no solution

8 0
3 years ago
Read 2 more answers
Theresa’s parents, Cal and Julia, are getting phones that each have 64 gigabytes of storage. How many bits of storage come with
r-ruslan [8.4K]
32 
32 x 10
hope it helps

8 0
3 years ago
Read 2 more answers
(02.01 LC)
ExtremeBDS [4]

Answer:

Yes it is a function

Step-by-step explanation:

We have to check the ordered pairs to find out if given relation is a function or not.

In an ordered pair, the first element represents the input and the second element represents the output.

The set of inputs is domain and output is range.

For a relation to be function, there should be no repetition in domain i.e there should be unique pairs of input and output.

In the given relation, the domain is {3,5,-1,-2}.

No element is repeated hence it is a function ..

7 0
3 years ago
What is the quadratic formula?
Ne4ueva [31]

Answer:

I guess u want to know the quadratic function and its formula is, AX*2 + BX + C. Otherwhise you want to know how to get 0 and get the Xm if you want to know how to solve a quadratic. That formula is,  (-B +/- \sqrt{x} (B*2 - 4AC) ) . 1/2 (where A, B and C are the number at the original function).

Step-by-step explanation:

6 0
3 years ago
Vector u has initial point at (3, 9) and terminal point at (–7, 5). Vector v has initial point at (1, –4) and terminal point at
spin [16.1K]

Answer:

⟨-5, -1⟩

Step-by-step explanation:

Vector:

A vector is given by its endpoint subtracted by its initial point.

Vector u has initial point at (3, 9) and terminal point at (–7, 5)

Then

u = (-7, 5) - (3,9) = (-7 - 3, 5 - 9) = (-10,-4)

Vector v has initial point at (1, –4) and terminal point at (6, –1).

Then

v = (6,-1) - (1,-4) = (6-1,-1-(-4)) = (5,3)

What is u + v in component form?

u + v = (-10,-4) + (5,3) = (-10+5,-4+3) = (-5,-1)

⟨-5, -1⟩ is the answer.

3 0
3 years ago
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