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Lera25 [3.4K]
3 years ago
15

B is between A and C if AB=3x+4 and BC= 4x-1 and AC=8x+15, find BC

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

BC = \begin{vmatrix}- 49\end{vmatrix} = 49   unit

Step-by-step explanation:

The given as AC is a line with B as its mid point

AND ,

AC = 8x + 15

AB = 3x + 4

BC = 4x - 1

∵ AC = AB + BC

Or, 8x + 15 = ( 3x + 4 ) + ( 4x - 1)

Or, 8x + 15 = 7x + 3

Or, 8x - 7x = 3 - 15

∴             x = - 12

SO,        AC = 8x + 15 = 8 × ( - 12) + 15 = - 81

Or,          AC =   \begin{vmatrix}- 81\end{vmatrix} = 81 unit

And       ,AB = 3x + 4 = 3 × ( - 12) + 4 = - 32

Or         ,AB =  \begin{vmatrix}- 32\end{vmatrix} = 32 unit

Similarly BC = 4x - 1 = 4 × ( - 12) - 1 = - 49

Or,         BC = \begin{vmatrix}- 49\end{vmatrix} = 49 unit Answer

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3 years ago
A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e
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Answer:

\frac{r(1-\sqrt{2})}{-3}

Step-by-step explanation:

Volume of cone = \frac{1}{3} \pi r^{2} h

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

                          = \frac{1}{3} \pi r^{2} \times r

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Since r = h

So, area = \pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}

              = \pi r^{2} +\pi\times r \times \sqrt{2r^2}

              = \pi r^{2} +\pi\times r^2 \times \sqrt{2}

Ratio of volume of cone to its surface area including base :

\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}

\frac{\frac{1}{3}r}{1+\sqrt{2}}

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Rationalizing

\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}

\frac{r(1-\sqrt{2})}{-3}

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is \frac{r(1-\sqrt{2})}{-3}

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Answer:

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2) 3 π/4 is in the third quadrant, so the reference angle is π - 3 π/4 = π/4

3) sin (π/4) = sin (3π/4) = (√2) / 2

4) csc (π/4) =  1 / [ sin (3π/4) ] = 1 / [ (√2) / 2 ] = 2 / (√2) = √2, which shows the validity of the statement and csc(19π/4) = √2.
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