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3 years ago

6

Given AJKL: AXYZ, find x.

1 answer:

Reil [10]3 years ago

8 0

Answer:

<u>6</u><u> </u> = <u>9</u>

8 x

we cross multiply 6× x = 6x 9×8 = 72

<u>6x</u><u> </u>= <u>72</u>

6 6

x = 12

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Soloha48 [4]

Total surface area equals area of 5 rectangular faces + 2 pentagonal faces

Area of 5 rectangular faces

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Area of 2 pentagonal faces can be found by drawing 5 isosceles triangles with height as shown in the diagram and multiply the result by 2.

$=5 \times \frac{1}{2} \times 9 \times 6.2 \times 2km^2$

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$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

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Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

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Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

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$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

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$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

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