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Artemon [7]
3 years ago
7

Help!! I believe I need to find the L and W of browns rectangle to make macdonalds equation but I’m stuck there and need to answ

er the whole thing for 11

Mathematics
1 answer:
34kurt3 years ago
8 0

Answer:

W man's W life W parents W pet W grand parents

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Yes, 2 is a rational number.

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Solve for the value of x:-<br> <img src="https://tex.z-dn.net/?f=4x-5%20%3D%203" id="TexFormula1" title="4x-5 = 3" alt="4x-5 = 3
enot [183]

4x-5 = 3

4x = 8

x = 8÷4

x = 2

5 0
2 years ago
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3 years ago
If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta
Karo-lina-s [1.5K]
\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad  &#10;\begin{array}{llll}&#10;\textit{now, hypotenuse is always positive}\\&#10;\textit{since it's just the radius}&#10;\end{array}&#10;\\\\\\&#10;thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow  hypotenuse=c}

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means,  -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

\bf \textit{Double Angle Identities}&#10;\\ \quad \\&#10;sin(2\theta)=2sin(\theta)cos(\theta)&#10;\\ \quad \\&#10;cos(2\theta)=&#10;\begin{cases}&#10;cos^2(\theta)-sin^2(\theta)\\&#10;\boxed{1-2sin^2(\theta)}\\&#10;2cos^2(\theta)-1&#10;\end{cases}&#10;\\ \quad \\&#10;tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\&#10;-----------------------------\\\\&#10;cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2&#10;\\\\\\&#10;cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}




\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}&#10;\\\\\\&#10;tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}&#10;\\\\\\&#10;tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}
6 0
3 years ago
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