4x + 2y = 6, -x +2y = -4

Mathematics

2 answers:

Delvig [45]3 years ago

5 0

Answer:

Step-by-step explanation:

No solution because the variable canceled it’s self

pochemuha3 years ago

3 0

The variable canceled itself already

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Zachary invested $130 in an account paying an interest rate of 3.6% compounded daily. Assuming no deposits or withdrawals are ma

TiliK225 [7]

Answer:

$193

Step-by-step explanation:

8 0

3 years ago

Use the squared identities to simplify 2sin^2xcos^2x

Rainbow [258]

$\bf sin^2({{ \theta}})=\cfrac{1-cos(2{{ \theta}})}{2}\qquad \qquad 

cos^2({{ \theta}})=\cfrac{1+cos(2{{ \theta}})}{2}\\
-----------------------------\\
thus
\\\\
2sin^2(x)cos^2(x)\implies \boxed{2}\cdot \cfrac{1-cos(2{{ x}})}{\boxed{2}}\cdot \cfrac{1+cos(2{{ x}})}{2}
\\\\
\cfrac{1-cos(2{{ x}})}{1}\cdot \cfrac{1+cos(2{{ x}})}{2}\implies 
\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\\
-----------------------------\\
\textit{now recall}\implies \textit{difference of squares}
\\ \quad \\
$

$\bf (a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\
-----------------------------\\
thus
\\\\
\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\implies \cfrac{1^2-cos^2(x)}{2}
\\\\
 \cfrac{1-cos^2(x)}{2}\impliedby 
\begin{array}{llll}
\textit{pretty sure you know what that numerator is}\\
\textit{check your pythagorean identities}
\end{array}$