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nikdorinn [45]
3 years ago
13

Hi can you please help me and explain step by step if possible thank you

Mathematics
1 answer:
zaharov [31]3 years ago
3 0

Answer:  see image

<u>Step-by-step explanation:</u>

Count how many spaces the vertex is away from the line x = -1.  Then place that vertex the same distance on the other side of the line x = -1.

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Answer? with solution please. Will automatically report kung wala:)
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B

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Communitive only works with multiplication and addition. It states that if you have something adding or subtracting something switched around, they are the same. B has that so the answer is B

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THE DIMENSIONS OF A CUBED ARE 4X C.M,6X C.M AND 10X C.M
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A. 248 square CM

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The perimeter of a rectangular carpet is 32 ft. Its length of the carpet is 6 ft. What is its width, w?
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Please help?
Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0
3 years ago
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