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To check our result, we note that if $x^2 + y^2 = 1$ then $\begin{pmatrix} X \\ Y \end{pmatrix}$ satisfy the equation [select all that apply] $\boxed{\text{✓}}\left(\frac{X}{3} + \frac{Y}{3}\right)^2 + \left(\frac{-2X}{3} + \frac{Y}{3}\right)^2 = 1$ $\boxed{\text{✓}}5X^2 - 2XY + 2Y^2 = 9$ $\Box(X - Y)^2 + (2X + Y)^2 = 1$ $\Box5X^2 + 2XY + 2Y^2 = 1$.

Name: to check our result we note that if x2 y2 1 then beginpmatrix x y endpmatrix satisfy the equation select all that apply boxedtextleftfracx3 fracy3right2 leftfrac 2x3 fracy3right2 1 boxedtext 31108
Uploaded: 2020-07-10T09:09:50-08:00
Duration: 2 min 22 s
Channel: Ankit Gupta
Description: to check our result we note that if x2 y2 1 then beginpmatrix x y endpmatrix satisfy the equation select all that apply boxedtextleftfracx3 fracy3right2 leftfrac 2x3 fracy3right2 1 boxedtext 31108

          To check our result, we note that if $x^2 + y^2 = 1$ then $\begin{pmatrix} X \\ Y \end{pmatrix}$ satisfy the
equation [select all that apply]

$\boxed{\text{✓}}\left(\frac{X}{3} + \frac{Y}{3}\right)^2 + \left(\frac{-2X}{3} + \frac{Y}{3}\right)^2 = 1$

$\boxed{\text{✓}}5X^2 - 2XY + 2Y^2 = 9$

$\Box(X - Y)^2 + (2X + Y)^2 = 1$

$\Box5X^2 + 2XY + 2Y^2 = 1$.

$to check our result we note that if x2 y2 1 then beginpmatrix x y endpmatrix satisfy the equation select all that apply boxedtextleftfracx3 fracy3right2 leftfrac 2x3 fracy3right2 1 boxedtext 31108$

Added by Gloria C.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Ankit Gupta

07/10/2020

Step 1

However, the problem statement uses $x$ and $y$ in the condition $x^2 + y^2 = 1$, but then uses $X$ and $Y$ in the vector and the equations. It is highly probable that $X$ and $Y$ are meant to be the same as $x$ and $y$. So, we will assume that $X^2 + Y^2 = Show more…

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To check our result, we note that if $x^2 + y^2 = 1$ then $\begin{pmatrix} X \\ Y \end{pmatrix}$ satisfy the equation [select all that apply] $\boxed{\text{✓}}\left(\frac{X}{3} + \frac{Y}{3}\right)^2 + \left(\frac{-2X}{3} + \frac{Y}{3}\right)^2 = 1$ $\boxed{\text{✓}}5X^2 - 2XY + 2Y^2 = 9$ $\Box(X - Y)^2 + (2X + Y)^2 = 1$ $\Box5X^2 + 2XY + 2Y^2 = 1$.