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2 years ago

9

Which equation can be simplified to find the inverse of y=5x^2+10

1 answer:

vodka [1.7K]2 years ago

6 0

F^-1(x) should be the answer.

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Solve the equation. Graph the solution(s) if possible.<br><br> ∣x−3∣+4=9

Mazyrski [523]

Answer:

x=8

x=-2

Step-by-step explanation:

∣x−3∣+4=9

= ∣x−3∣=5

<u>Positive term: (x-3)</u>

(x-3) = 5

Rearrange and Add up

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<u>Negative term: -(x-3)</u>

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Step-by-step explanation:

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solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything

vova2212 [387]

a.

$XA^{-1}=A^3$

$(XA^{-1})A=A^3A$

$X(A^{-1}A)=A^4$

$X=A^4$

b.

$AXB=(BA)^2$

$A^{-1}(AXB)B^{-1}=A^{-1}(BA)^2B^{-1}$

$(A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}$

$X=A^{-1}(BA)^2B^{-1}$

c.

$(A^{-1}X)^{-1}=(AB^{-1})^{-1}(AB^2)$

$X^{-1}A=(BA^{-1})(AB^2)$

$X^{-1}A=B(A^{-1}A)B^2$

$X^{-1}A=B^3$

$(XX^{-1})A=XB^3$

$XB^3=A$

$X(B^3(B^3)^{-1})=A(B^3)^{-1}$

$X=A(B^3)^{-1}$

d. Not totally sure what the equation is supposed to be, but I guess it's

$ABXA^{-1}B^{-1}=A$

$ABX(BA)^{-1}=A$

$((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)$

$X=(AB)^{-1}A(BA)$

$X=(B^{-1}A^{-1})A(BA)$

$X=B^{-1}(A^{-1}A)(BA)$

$X=(B^{-1}B)A$

$X=A$

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2 years ago