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antiseptic1488 [7]
4 years ago
7

Convert to polar form y=3x^2

Mathematics
2 answers:
cluponka [151]4 years ago
7 0

<span>r=2sin0-3cos0</span>

explain<span>using the formula that links Cartesian to Polar coordinates.<span>∙y=r<span>sinθ</span></span><span>∙x=r<span>cosθ</span></span>then : <span>r<span>sinθ</span>=3r<span>cosθ</span>+2</span>and <span>r<span>sinθ</span>−3r<span>cosθ</span>=2</span>hence <span>r<span>(<span>sinθ</span>−3<span>cosθ</span>)</span>=2</span></span>

<span>⇒r=<span><span>2<span><span>sinθ</span>−3<span>cosθ</span></span></span></span></span>

Levart [38]4 years ago
6 0

<span>r = <span>2/(<span><span>sinθ</span>−3<span>cosθ)</span></span></span></span>

Explanation:<span>y = r sin θ
<span>x = r <span>cos θ</span></span>Then: <span>r <span>sin θ</span>=3r <span>cos θ </span>+ 2</span>and <span>r <span>sin θ </span>−3r <span>cos θ </span>= 2</span>thus <span>r <span>(<span>sin θ</span>−3 <span>cos θ</span>) </span>=2</span></span>

<span>⇒r =<span><span>2/(<span><span>sin θ</span>−3 <span>cos θ)</span></span></span></span></span>

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2 years ago
How to factor ax2 bx c?
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In the same way as you could factor trinomials on the form of

<span><span><span>x2</span>+bx+c</span><span><span>x2</span>+bx+c</span></span>

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<span><span>a<span>x2</span>+bx+c</span><span>a<span>x2</span>+bx+c</span></span>

If a is positive then you just proceed in the same way as you did previously except now

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FrozenT [24]

Answer and Step-by-step explanation:

Alot of these can be solved with the table up above the problems.

<h2>A, B, and C.</h2><h2></h2>

According to the table, every time a value has an exponent of 0, it would equal to 1. Since all A B and C values follow the 0 exponent, their value is 1.

<h2>D.</h2><h2 />

According to the table, when we have a negative exponent, the number gets turned into a fraction of \frac{1}{x^x}

Since our value is 9^-2, do as followed :

\frac{1}{9^2}

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[\frac{1}{81} ]

<h2 /><h2>E.</h2><h2 />

\frac{1}{x^{-5} }

This question is different, since we have to go backwards, we have to switch the negative exponent to a positive exponent and rid the fraction.

x^5

<h2>F.</h2><h2 />

5a^3b^{-2}

Since we have one negative exponent, follow the fraction rule but instead of having one as the numerator, have 5a^3.

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<h2>Fill in the table :</h2><h2 />

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Since -3 is in parenthesis, we can rid the negative along with the parenthesis :

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Follow the negative exponent rule yet again :

-3^{-4}=-\frac{1}{3^4} =-\frac{1}{3*3*3*3} =-\frac{1}{81}

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