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1 year ago

Solve by factoring 6p^2 =21p-6

Mathematics

1 answer:

uysha [10]1 year ago

7 0

Given,

The expression is:

$6p^2=21p-6$

By using the middle term splitting method,

$\begin{gathered} 6p^2-21p+6=0 \\ 2p^2-7p+2=0 \\ (x-(\frac{7+\sqrt{33}}{4}))(x+\frac{7+33}{4})=0 \end{gathered}$

Hence, the factor is (x-(7+sqrt(33)/4)) (x+(7+sqrt(33)/4)).

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<span>My estimate was 2,808 is that a reasonable estimate and why

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3 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

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$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
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\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
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I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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3 years ago

What is the fully factored form of the expression 16x^2 - 49y^2

podryga [215]

Answer:

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Step-by-step explanation:

Rewrite 16 as 4^2

= 4^2x^2 - 49y^2

Rewrite 49 as 7^2

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Apply the Exponent Rule Pt 1 ((a^m*b^m =(ab)^m))

= (4x)^2 - 7^2y^2

Apply the Exponent Rule Pt 2

= (4x)^2 - (7y)^2

Apply Difference of Squares Formula (( x^2-y^2 = (x + y)(x - y)

= (4x + 7y) (4x - 7y)

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3 years ago