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Lena [83]
3 years ago
13

Help me i need to submit to my teacher by 8pm​

Mathematics
1 answer:
Vladimir [108]3 years ago
7 0

Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.

Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.

A/2 = area(sector PTQ) - area(triangle PUT)

The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so

TU = (7/2)√3

area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3

area(sector PTQ) = (angle TQP / 360°) πr^2

We know angle TQP is 60° because TQP is equilateral.  r=7.

area(sector PTQ) = (60°/360°) π (7²) = 49π/6

Putting it together,

A/2 = area(sector PTQ) - area(triangle PUT)

A = 2(49π/6 -  (49/8)√3)

A = 49(π/3 - √3/4) square cm

I hate ruining a nice exact answer with an approximation, but they seem to be asking.

A ≈ 30.095057615914535

Check:

I'm not sure how to check it.  I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.

Now the real area we seek is the big square PQRS minus A, so

area = 7² - 30.095057615914535 = 18.904942384086 sq cm

They want square meters for some reason; we scale by (1/100)²

Answer: 0.00189 square meters

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AlladinOne [14]

Answer:

The mode are: 7, 8

Step-by-step explanation:

Given

7, 11, 8, 7, 8, 7, 8, 11, 8, 10, 7

Required

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We start by arranging the given data in ascending order

The ordered data are:

7,7,7,7, 8,8,8,8, 10, 11, 11

From the above data.

7 has a frequency of 4

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10 has a frequency of 1

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The data with the highest frequency is the mode.

We can see that 7 and 8 both have frequencies of 4

<em>Hence, the mode are: 7, 8</em>

8 0
3 years ago
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77julia77 [94]

For  a height of 4.3m and bases of 9.2 m and 3.2m, the area of the trapezoid  is mathematically given as

A=26.7

<h3>What is the area of the trapezoid?</h3>

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Helppppp pls im stuck on thi one problem :(
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First, find the simplified ratio of the given ratio:

Number of pirates: 22

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For every ship there is, there will be 11 pirates.

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Andrew
PilotLPTM [1.2K]
Let's say Bailey takes "b" days to paint it.

and Andrew takes "a" days to paint the same house.

now, Andrew is 6 times faster than Bailey, therefore, if Andrew takes "a" days to do it, Bailey takes then "6a" days, or b = 6a.

now, the year they worked together, they finished it in 7 days.

so, after 1 day then, they have only done 1/7 of the whole work.

and Andrew for one day, has done 1/a of the house, whilst Bailey has done 1/b of the house or 1/(6a).

\bf \stackrel{\textit{Andrew's rate}}{\cfrac{1}{a}}+\stackrel{\textit{Bailey's rate}}{\cfrac{1}{b}}=\stackrel{\textit{1 day of work}}{\cfrac{1}{7}}&#10;\\\\\\&#10;\cfrac{1}{a}+\cfrac{1}{6a}=\cfrac{1}{7}\impliedby &#10;\begin{array}{llll}&#10;\textit{let's multiply all by }\stackrel{LCD}{42a}\textit{ to toss the}\\&#10;denominators&#10;\end{array}

\bf 42a\left( \cfrac{1}{a}+\cfrac{1}{6a} \right)=42a\left( \cfrac{1}{7} \right)\implies 42+7=6a\implies \cfrac{49}{6}=a&#10;\\\\\\&#10;\stackrel{days}{8\frac{1}{6}}=a&#10;\\\\\\&#10;\textit{how many days will it take Bailey then?}\quad b=6a&#10;\\\\\\&#10;b=6\cdot \cfrac{49}{6}\implies b=\stackrel{days}{49}
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