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3 years ago

11

What is the inverse of y=2x squared -4

1 answer:

Bumek [7]3 years ago

6 0

$y=2x^2-4\ \ \ |+4\\\\2x^2=y+4\ \ \ |:2\\\\x^2=\dfrac{y+4}{2}\\\\x=\sqrt{\dfrac{y+4}{2}}$

$y^{-1}=\sqrt{\dfrac{x+4}{2}}=\dfrac{\sqrt{x+4}}{\sqrt2}=\dfrac{\sqrt{2(x+4)}}{2}=\dfrac{\sqrt{2x+8}}{2}$

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Please help! I would also like an explanation on how you find the answer if possible.

pentagon [3]

check the picture below.

3 0

3 years ago

What is the solution to the equation StartFraction 3 Over m + 3 EndFraction minus StartFraction m Over 3 minus m EndFraction = S

Oksanka [162]

Answer:

No solution.

Step-by-step explanation:

Given fraction is:

$\dfrac{3}{m+3} -\dfrac{m}{3-m} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3}{m+3} +\dfrac{m}{m-3} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3\times (m-3)+ m(m+3)}{(m+3)(m-3)} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3 m-9+ m^2+3m}{m^2-9} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{m^2-9+ 6m}{m^2-9} = \dfrac{m^2+9}{m^2-9}\\If\ m^2-9 \neq 0\ or\ m \neq 3\\\Rightarrow m^2-9+ 6m = m^2+9\\\Rightarrow 6m = 18\\\Rightarrow m = 3$

<u>Formula used:</u>

<u></u> $(a+b)(a-b) = a^{2} -b^{2}$ <u></u>

<u></u>

But, the above equation was solved at the condition that:

$m\neq 3$

So, there is no solution for the given equation.

3 0

3 years ago

Read 2 more answers

Look at picture and select all that appy

Lyrx [107]

Answer:

the first one is scalne the second is isoscles and the third is equailateral

Step-by-step explanation:

scalne- all sides diff lengths

isoscles- two sides same length (remeber the two s in iSoScles)

equaliteral- all sides same length

8 0

3 years ago

What’s the differential equation of

worty [1.4K]

Answer:

$y=\frac{1}{x^{2}-6x+13 }$

Step-by-step explanation:

We have given,

$\frac{dy}{dx}=y^{2}(6-2x)$

and initial condition $x=3,\ y=\frac{1}{4}$

Now,

$\frac{dy}{dx}=y^{2}(6-2x)$

Rearranging the variables, we get

$\frac{dy}{y^{2} }=(6-2x).dx$

Applying integration both sides, we get

$\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx$

⇒ $\frac{-1}{y} =6x-\frac{2x^{2} }{2}$

⇒ $\frac{-1}{y}=6x-x^{2} +C$

Putting the initial condition (i.e., $x=3,\ y=\frac{1}{4}$ ), we get

⇒ $-4=6\times3-(3)^{2}+C$

⇒ $-4=18-9+C$

∴ $C=-13$

We have, $\frac{-1}{y}=6x-x^{2} +C$

now putting the value of $C$ in above equation, we get

⇒ $\frac{-1}{y}=6x-x^{2} -13$

⇒ $\frac{1}{y}=-6x+x^{2} +13$

$y=\frac{1}{x^{2}-6x+13 }$

5 0

3 years ago

Thankyou guys for all your help….I need help with one more plsss<br><br> Round to the nearest tenth

avanturin [10]

I don't understand your question

3 0

2 years ago

Read 2 more answers