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svetlana [45]
3 years ago
12

I answered the question 3x-y=5 but is there a different way to answer this?

Mathematics
2 answers:
Svetach [21]3 years ago
7 0

Answer and Step-by-step explanation:

Yes (technically),  there is a different way to answer the equation 3x - y = 5. But, the first part of how you described it is a bit off.

Here is how you would explain it:

What you would do is Add y to both sides, then subtract 5 from both sides.

3x = y + 5

3x - 5 = y (aka y  = 3x - 5)

The other way to solve:

3x - y = 5

Subtract 3x from both sides.

-y = -3x + 5

Now divide both sides by -1.

-1(-y) = -1(-3x + 5)

y = 3x - 5

If you want to solve for x, you would:

Add y to both sides of the equation.

3x = 5 + y

Then divide each side by 3.

x = \frac{1}{3}y + \frac{5}{3}

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

Svet_ta [14]3 years ago
3 0

Step-by-step explanation:

Hi,

I agree with what you did. You could also add y on both sides and get 3x = y + 5 and then divided by three to get...

x = 5 + y /3

(the 5 and y are being divided by 3)

But that's just complicated, it works, but it's complicated. What you did is correct and what I gave you is the value of x, but I'm 100 percent sure yours is correct.

Hope this helps

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Prove the following
fomenos

Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
  • sin (270°-θ)  =  -cos θ
  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
  • cot (270°+θ)  =  -tan θ
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Step-by-step explanation:

Hey there!

Given;

The points is; (-1 , 2) and slope is 4.

<em>Note:</em><em> </em><em>Use </em><em>one-point</em><em> formula</em><em>.</em>

<em>(y - y1) = m(x - x1)</em>

<em>~</em><em> </em><em>Put </em><em>all</em><em> values</em><em>.</em>

(y - 2) = 4(x + 1)

~ Simplify it.

(y - 2) = 4x + 4

4x - y + 6 = 0

Therefore, the required equation is; 4x-y+6=0.

<h3><em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h3>

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Answer:

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4 years ago
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Answer: 4/6

1. Subtract 8/6 by 4/6.

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<em>That's why 4 + 4 = 8 is the same as 4/6 + 4/6.</em>

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