All Answers 356032 Answers

0.33333333........ = 1/3

3/4 foot per week. The lake is dropping an average of 3/4 of a foot per week. We don't know compared "to its full level", because you don't have the depth of the lake when full.

Selecciona el cuadro y la ecuación que represente las funciones.

Cuadro 1:

x12345

y246810

Cuadro 2:

x12145

y245810

Ecuación 1:

y = 13x + 6y = 13x + 6

Ecuación 2:

x2 + y2 = 1x2 + y2 = 1

hey bud its your dad ricky. I am so tired of you being this dumb get a life 

well if you start with one cat and it gives birth to 100 cats thats not scientificaly possibble but that means you have 101 cats and then you take 101 and cut it in half because half of them were brutaly murdered. answer 50.5 cats be

if 5.3 cm is 120, what is 4

(120/5.3) x 4 =90.566

well if you start with one cat and it gives birth to 100 cats thats not scientificaly possibble but that means you have 101 cats and then you take 101 and cut it in half because half of them were brutaly murdered. answer 50.5 cats because the wold ate half of one of the cats. Good question.

12x20=240

what is the 8% of 54000

8/100 x 54,000 =4,320

Factor the following:
x^4-7 x^2-144

x^4-7 x^2-144 = (x^2)^2-7 x^2-144:
(x^2)^2-7 x^2-144

The factors of -144 that sum to -7 are 9 and -16. So, (x^2)^2-7 x^2-144 = (x^2+9) (x^2-16):
(x^2+9) (x^2-16)

x^2-16 = x^2-4^2:
(x^2+9) (x^2-4^2)

Factor the difference of two squares. x^2-4^2 = (x-4) (x+4):
Answer: |(x-4)  (x+4)  (x^2+9)

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i dont know

4320

dont know

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I'm good but pls only post math related questions!

Three equal circles with radius r are drawn as shown,
each with its centre on the circumference of the other two circles.
A, B and C are the centres of the three circles.
Prove that an expression for the area of the shaded region is:

\(A=\frac{r^2}{2}\left(\pi -\sqrt{3}\right)\)

\(\begin{array}{|rcll|} \hline A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\\\ h_{\triangle ABC}^2 + (\frac{r}{2})^2 &=& r^2 \\ h_{\triangle ABC}^2 + \frac{r^2}{4} &=& r^2 \\ h_{\triangle ABC}^2 = r^2 - \frac{r^2}{4} \\ h_{\triangle ABC}^2 = \frac34 \cdot r^2 \\ h_{\triangle ABC} = \frac{\sqrt{3}}{2} \cdot r \\\\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot \frac{\sqrt{3}}{2} \cdot r \\ \mathbf{A_{\triangle ABC} }&\mathbf{=}& \mathbf{\frac{r^2}{4} \cdot \sqrt{3} }\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A_{Arc} &=& \pi r^2 \cdot \frac{60^{\circ}} {360^{\circ}} \\ A_{Arc} &=& \frac{\pi r^2} {6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_{\text{shaded region}} &=& 3\cdot (A_{Arc}-A_{\triangle ABC})+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-3\cdot A_{\triangle ABC}+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-2\cdot A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot \frac{\pi r^2} {6}-2\cdot \frac{r^2}{4} \cdot \sqrt{3} \\ A_{\text{shaded region}} &=& \frac{\pi r^2} {2}- \frac{r^2}{2} \cdot \sqrt{3} \\ \mathbf{A_{\text{shaded region}} }&\mathbf{=}& \mathbf{\frac{r^2}{2} \cdot ( \pi- \sqrt{3} )} \\ \hline \end{array}\)

Hi Kreyn,

Three equal circles with radius r are drawn as shown, each with its centre on the circumference of the other two circles. A, B and C are the centres of the three circles. Prove that an expression for the area of the shaded region is:

\(A=\frac{r^2}{2}\left(\pi -\sqrt{3}\right)\)

Now AB, AC and BC are all radii, so they are all equal, so ABC is an equilateral triangles and all the angles are 60 degrees.

\(\boxed{\text{Area of triangle ABC=}\frac{1}{2}absinC }\\ \\~\\ Area\;\;ABC=\frac{1}{2}r^2sin60^\circ\\ Area\;\;ABC=\frac{1}{2}r^2*\frac{\sqrt3}{2}=\frac{\sqrt3\;r^2}{4}\)

Now I want to know what the area of minor segment AB is on the circle centred at C

\(Segment\; area = \frac{60}{360}\pi r^2\;-\;\frac{\sqrt3\;r^2}{4}\\ Segment\; area = \frac{2}{12}\pi r^2\;-\;\frac{3\sqrt3\;r^2}{12}\\ Segment\; area = \frac{2\pi r^2-3\sqrt3\;r^2}{12}\\ so\\ Shaded \;area = 3* \frac{2\pi r^2-3\sqrt3\;r^2}{12} + \;\frac{\sqrt3\;r^2}{4}\\ Shaded\; area = \frac{2\pi r^2-3\sqrt3\;r^2+\sqrt3\;r^2}{4} \\ Shaded\; area = \frac{2\pi r^2-2\sqrt3\;r^2}{4} \\ Shaded\; area = \frac{\pi r^2-\sqrt3\;r^2}{2} \\ Shaded\; area = \frac{ r^2}{2}(\pi -\sqrt3\;) \\\)

8

Y'all idiots in Astro 1F03?

inb4 Couchman calls you idiots out on cheating off the internet

Numbers?????????