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2 years ago

Each leg of a 45°-45°-90° triangle measures 12 cm what is the hypoyenuse

1 answer:

5 0

To find the hypotenuse in a 45-45-90 triangle multiply the legs by $\sqrt{2}$
So...
The leg is 12 the hypotenuse is 12 $\sqrt{12}$

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$Day \ \ \ \ \ \ \ \ \ \ Period \ \ \ \ \ \ \ \ \ \ Demand \ \ \ \ \ \ \ \ \ \ 5-period moving average$

$1 \ \ \ \ \ \ \ \ \ \ \ Morning \ \ \ \ \ \ \ \ \ \ \ 37 \\\\$

$Afternoon \ \ \ \ \ \ \ \ \ \ \ 25 \\\\ Evening \ \ \ \ \ \ \ \ \ \ \ 9$

$2 \ \ \ \ \ \ \ \ \ \ \ Morning \ \ \ \ \ \ \ \ \ \ \48 \\\\$

$Afternoon\ \ \ \ \ \ \ \ \ \ \ 28 \\\\$

$Evening \ \ \ \ \ \ \ \ \ \ \ 14 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 29.4 \\\\$

$3 \ \ \ \ \ \ \ \ \ \ \ Morning \ \ \ \ \ \ \ \ \ \ \ 53 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 24.8\\\\$

$Afternoon \ \ \ \ \ \ \ \ \ \ \ 33 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30.4\\\\Evening \ \ \ \ \ \ \ \ \ \ \ 17 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35.2\\\\$

$4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Morning \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 29\\\\$

$Afternoon\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 29.25\\\\Evening \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34.33333\\\\$

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