Answer:
y = (x/(1-x))√(1-x²)
Step-by-step explanation:
The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:
x = r·cos(θ)
y = r·sin(θ)
x² +y² = r²
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r = sec(θ) -2cos(θ)
r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)
r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²
(x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations
x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor
y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)

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The attached graph shows the equivalence of the polar and rectangular forms.
Answer:
5. LCM of 7 and 14: <u> </u><u> </u><em><u>1</u></em><em><u>4</u></em><em><u>. </u></em>
multiples of 7: <u> </u><u> </u><u>7</u><u>,</u><u> </u><u>1</u><u>4</u><u> </u>
multiples of 14: <u> </u><u>1</u><u>4</u><u> </u>
LCM of 8 and 12: <u> </u><u> </u><em><u>2</u></em><em><u>4</u></em><em><u>. </u></em>
multiples of 8: <u> </u><u> </u><u>8</u><u>,</u><u> </u><u>1</u><u>6</u><u>,</u><u> </u><u>2</u><u>4</u><u> </u>
multiples of 12: <u> </u><u> </u><u>1</u><u>2</u><u>,</u><u> </u><u>2</u><u>4</u><u> </u>
Step-by-step explanation:

Answer:
P(4≤x≤7) = 2/3
Step-by-step explanation:
We'll begin by obtaining the sample space (S) i.e possible outcome of rolling both dice at the same time. This is illustrated below:
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
Adding the outcome together, the sample space (S) becomes:
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
Next, we shall obtain the event of 4≤x≤7. This is illustrated below:
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
Finally, we shall determine P(4≤x≤7). This can be obtained as follow:
Element in the sample space, n(S) = 36
Element in 4≤x≤7, n(4≤x≤7) = 24
Probability of 4≤x≤7, P(4≤x≤7) = ?
P(4≤x≤7) = n(4≤x≤7) / nS
P(4≤x≤7) = 24/36
P(4≤x≤7) = 2/3
the circle has a diameter of 12, thus its radius is half that, or 6.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=6 \end{cases}\implies A=\pi 6^2\implies A=36\pi \\\\\\ A \approx 113.0973355\implies A=\stackrel{\textit{rounded up}}{113.1}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D6%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%206%5E2%5Cimplies%20A%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20A%20%5Capprox%20113.0973355%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B113.1%7D)