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diamong [38]
3 years ago
13

Find S15 for the geometric series 72 + 12 + 2 + 1/3 +

Mathematics
2 answers:
Rama09 [41]3 years ago
8 0
Finish the equation. please and thank you

sweet-ann [11.9K]3 years ago
5 0
Hello,

u_{1}=72\\

u_{2}=\dfrac{72}{6}\\

u_{3}=\dfrac{72}{6^{2}}\\

u_{15}=\dfrac{72}{6^{14}}\\


s=72*(1+\dfrac{1}{6^{1}}+\dfrac{1}{6^{2}}+\dfrac{1}{6^{3}}++...+\dfrac{1}{6^{14}})\\
=72*(\dfrac{\frac{1}{6^{15}}-1}{\frac{1}{6}-1}})\\
=72*(\dfrac{470184984576-1}{78364164095*5})\\

=86,400...



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melamori03 [73]

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6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

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The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

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The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

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s=\frac{5+5+4}{2}=\frac{14}{2}=7

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Thus, the area of the quadrilateral is:

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