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2 years ago

- What is the area of the parallelogram below? 4in 2.8 in 9in

Mathematics

2 answers:

vaieri [72.5K]2 years ago

8 0

Answer:

100.8

I hope this helps:D

coldgirl [10]2 years ago

4 0

Answer:

100.8

Step-by-step explanation:

multiply 4 2.8 and 9 and you get 100.8

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A ball is thrown upward off of a 100 meter cliff with an initial velocity of

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108

I'm not positive but i belive this is correct

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3 years ago

a highway measures 58 and 1/2 miles long. There will be a speed limit sign posted every two one six miles. How many speed limit

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a highway measures 58 and 1/2 miles long.there be a speed limit sign posted every two one six miles.How many speed limit signs Will be on the highway 

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2 years ago

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.

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3 years ago

Which figure shows a reflection of pre-image DEFG over the x-axis?

Sergio [31]

The image that shows the reflection of DFEG over the x-axis is the FIRST option. The x-axis is the mirror line. The distance from the mirror line to the reflected shape equals to the distance from the original image to the mirror line.

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3 years ago

13.6z - 11 = -1.48. can u please show the work for it

serious [3.7K]

Step-by-step explanation:

13.6z - 11 = -1.48.

13.6z=-1.48+11

z=9.52/13.6=0.7

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2 years ago