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disa [49]
3 years ago
5

Simplify The Expression 2y - y + 4 + y y=3​

Mathematics
1 answer:
Brut [27]3 years ago
7 0

-0.5

Step-by-step explanation:

um bc it is the answer

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Point B has coordinates ​(1​,2​). The​ x-coordinate of point A is negative 8. The distance between point A and point B is 15 uni
olchik [2.2K]

Answer:

The point A will be  (-8,14) or (-8,-10).

Step-by-step explanation:

Point B has coordinates (1,2) and the x-coordinate of point A is - 8.

Let us assume that the coordinates of point A are (-8,k).

Now, given that the point A is 15 units apart from point B.

Therefore, from the distance formula, we can write that  

\sqrt{(1 - ( -8))^{2} + (2 - k)^{2}} = 15

Now,squaring both sides, we get  

(1 - ( -8))^{2} + (2 - k)^{2} = 225

⇒ (2 - k)^{2} = 225 - 9^{2} = 144

⇒ 2 - k = ± 12

⇒ k = 14 or -10.

Therefore, the point A will be  (-8,14) or (-8,-10). (Answer)

We know that the distance between two points on the coordinate plane (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by  

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}.

7 0
3 years ago
Identify the correct logical reason for the next step in solving this equation
Musya8 [376]

Answer:

Distributive property of equality

Step-by-step explanation:

The first thing to do in an equation is to distribute.

6 0
2 years ago
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kaheart [24]
Communitive Property of Addition
4 0
3 years ago
Will mark brainliest
Varvara68 [4.7K]

Answer: I don’t think u gave enough info

Step-by-step explanation:

3 0
3 years ago
Given that f(a+b)= f(a) + f(b) and f(x) is always positive, what os the value of f(0)?
FrozenT [24]
This is a strange question, and f(x) may not even exist. Why do I say that? Well..

[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).

[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!

Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.

To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
4 0
3 years ago
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