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3 years ago

What is the base plan for the set if stacked cubes???plz help me im stuck and im in a hurry !!!!!

Mathematics

2 answers:

Sladkaya [172]3 years ago

4 0

I think it is the 3 one from the left side

lawyer [7]3 years ago

3 0

Answer: The answer is the first optional figure, attached the image.

Step-by-step explanation: We are given to choose the correct base plan for the set of stacked cubes in the figure.

We see that the base of the stacked cubes matches with the first optional figure, where there are 2 cubes in the front, 2 cubes perpendicular to the front cubes and then 1 more cube placed perpendicular to them in the left.

Thus, the correct image is the first image, attached herewith.

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aev [14]

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A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.

The given equation consisting of variable , m is

$\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})$

None of the two solution

$m=\frac{3}{2}\times(-1 +\sqrt{2}) and , m=\frac{3}{2}\times(-1 -\sqrt{2})$ , is extraneous.

Here, L.H.S= R.H.S

Option A: 0→ extraneous

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3 years ago

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