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mash [69]
3 years ago
13

Please help!!!

Mathematics
1 answer:
Lerok [7]3 years ago
5 0

The sum of x^2−3xy−y^2 and 2x^2+5xy−4y^2:

x^2−3xy−y^2 + 2x^2+5xy−4y^2

= 3x^2 + 2xy - 5y^2


From 6x^2−7xy+8y^2 subtract the sum of x^2−3xy−y^2 and 2x^2+5xy−4y^2:

6x^2−7xy+8y^2 - (3x^2 + 2xy - 5y^2)

= 6x^2−7xy+8y^2 - 3x^2 - 2xy + 5y^2

= 3x^2−9xy + 13y^2


Answer

3x^2−9xy + 13y^2

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katrin2010 [14]

\large\boxed{x=-\frac{7}{2}}

To solve for x, we need to isolate it on one side of the equation.

The most important part of this is knowing that whatever we do to one side of the equation, we must also do to the other.

Subtract 32 from both sides of the equation.

\begin{aligned}32-32-6x&=53-32\\-6x&=21\end{aligned}

Divide both sides of the equation by -6.

\begin{aligned}\frac{-6x}{-6}&=\frac{21}{-6}\\x&=\boxed{-\frac{7}{2}}\end{aligned}

7 0
2 years ago
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Alekssandra [29.7K]

Answer: 1 out of 4

Step-by-step explanation:

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4 years ago
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MAXImum [283]

Observe that

\sin\left(\dfrac\pi6-v\right)=\sin\dfrac\pi6\cos v-\cos\dfrac\pi6\sin v=\dfrac12\cos v-\dfrac{\sqrt3}2\sin v

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-\sqrt3\sin v+\cos v=\sqrt3\implies\dfrac12\cos v-\dfrac{\sqrt3}2\sin v=\dfrac{\sqrt3}2

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Next,

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where n is any integer. Then

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3 years ago
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Answer:

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Step-by-step explanation:

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Naddik [55]
Use order of operations. 

7 0
3 years ago
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