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OlgaM077 [116]
3 years ago
14

What is the first step to solve -2x+5x=9​

Mathematics
2 answers:
Mashutka [201]3 years ago
6 0

Answer:

Step-by-step explanation:

Hello,

You need to group the terms in x

-2 \times x + \times x = (-2+5) \times x = 3 x

And then, you can divide by 3 to find the solution

-2x+5x=3x=9x=9/3=3

Hope this helps

Thepotemich [5.8K]3 years ago
5 0

Answer:

add the x's

You have to combine them

you have to subtract the -2x to the 5x

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We want a in terms of b and c:     <span>3a+4b=c

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Divide all terms by 3:

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Another expression for the given rational function is x =r^\frac{1}{2}s^\frac{1}{2}

<h3>Indices expression</h3>

Given the rational function expressed as:

x = √rs

We can reconstruct the expression by writing it using exponents as shown;

x =\sqrt{rs}=(rs)^{\frac{1}{2} }

Separate to have:

x =r^\frac{1}{2}s^\frac{1}{2}

Hence another expression for the given rational function is x =r^\frac{1}{2}s^\frac{1}{2}

learn more on rational function here: brainly.com/question/1851758

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klio [65]

Answer:   \bold{-\dfrac{7\sqrt2}{10}}

<u>Step-by-step explanation:</u>

It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0).  Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ cos\ \dfrac{\theta}{2}

It is given that sin θ = -\dfrac{7}{25}  ⇒  y = -7 & hyp = 25

Use Pythagorean Theorem to find "x":

x² + y² = hyp²

x² + (-7)² = 25²

x² + 49 = 625

x²         = 576

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Use the "x" and "hyp" values to find cos θ:

cos\ \theta=\dfrac{x}{hyp}=\dfrac{24}{25}    


Lastly, input cos θ into the half angle formula:

cos\bigg(\dfrac{\theta}{2}\bigg)=\pm \sqrt{\dfrac{1+cos\ \theta}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{1+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{25}{25}+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{49}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{49}{50}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\\\\\\.\qquad \quad =\pm \dfrac{7\sqrt2}{10}

Reminder: We previously determined that the half-angle will be negative.

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