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Lina20 [59]
3 years ago
12

Find the area of each shade region (can you do both for me please)

Mathematics
2 answers:
MrMuchimi3 years ago
8 0

2a. Answer:   100

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

Area_{\ shaded\ region}=Area_{\ parallelogram}-Area_{\ triangle}\\\\Area_{\ parallelogram}=base \times height\\\\.\qquad \qquad \qquad \qquad =14\times 10\\\\.\qquad \qquad \qquad \qquad =140\\\\Area_{\ triangle}=\dfrac{base\times height}{2}\\\\\\.\qquad \qquad \quad =\dfrac{8\times 10}{2}\\\\\\.\qquad \qquad \quad =40\\\\Area_{\ shaded\ region}=140-40\\\\.\qquad \qquad \qquad \qquad =100

*******************************************************************************

2b. Answer:   114

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

Area_{\ shaded\ region}=Area_{\ trapezoid}-Area_{\ kite}\\\\Area_{\ trapezoid}=\dfrac{base_1+base_2}{2}\times height\\\\\\.\qquad \qquad \qquad =\dfrac{11+18}{2}\times (5+7)\\\\\\.\qquad \qquad \qquad =\dfrac{29\times 12}{2}\\\\\\.\qquad \qquad \qquad =174\\\\\\Area_{\ kite}=\dfrac{diagonal_1\times diagonal_2}{2}\\\\\\.\qquad \qquad=\dfrac{(5+5)(5+7)}{2}\\\\\\.\qquad \qquad=\dfrac{10\times 12}{2}\\\\\\.\qquad \qquad=60

Area_{\ shaded\ region}=174-60\\\\.\qquad \qquad \qquad \qquad =114

grandymaker [24]3 years ago
3 0
*FORMULA*

Qu1.

area \: of \: a \: triangle = \frac{1}{2} base \: \times height

Height=10

Base=8

= \frac{1}{2} \times 8 \times 10

= 40

Qu2.
Area of a kite=half×product of digonals
Step by step solution
Summation of 1st diagonal =5+5
=10
Summation of 2nd diagonal =7+5
=12
Product of the diagonals
=10×12
=120
Area=1÷2×120
=60
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Assume that thermometer readings are normally distributed with a mean of degrees and a standard deviation of 1.00degrees C. A th
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Answer:

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3 years ago
16 The perimeter of the pentagon is equal to the perimeter of the square.
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7x+2

Step-by-step explanation:

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2 years ago
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Area = 3.14 x 0.7^{2}

Area = 3.14 x 0.49

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round to the nearest hundredth:

1.5386 = 1.54

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