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allsm [11]
3 years ago
6

Given: 5(x - 2) = 2x - 4 Prove: What’s the The proof ?

Mathematics
1 answer:
Elena L [17]3 years ago
8 0

Answer:

Step-by-step explanation:

its not right equation

because

5X - 10 ≠ 2X - 4

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What is the answer to this question?
melisa1 [442]

Answer: I think it is negative.


Step-by-step explanation: When you add a negative to a negative you still have a negative.


7 0
4 years ago
The point (6, - 5) is reflected across the y axis what is the new point
jonny [76]

Answer:

  (-6, -5)

Step-by-step explanation:

Reflection across the y-axis leaves the point on the same horizontal line, but with the sign of its x-coordinate changed.

  (x, y) ⇒ (-x, y) . . . . . reflection across the y-axis

  (6, -5) ⇒ (-6, -5)

The image point is (-6, -5).

8 0
2 years ago
jennifer is shopping with her mother they pay $2 per pound for tomatoes at the vegetable stand. find the constant of proportiona
Tresset [83]
One the first problem on finding the constant of <span> proportionality in this situation the answer is 2. Below is the solution:

constant = unit rate
unit rate = $2/ poung 
K = 2

For the second question, the equation is the below:

T = 2p or y = 2x</span>
3 0
3 years ago
Read 2 more answers
I thought it was 25/100 buts it’s wrong
viva [34]

Answer:

Step-by-step explanation:

a to b represents a quarter of the line segment in order to get to %100 we need to multiply 2in by 4.

2x4=8

4 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
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