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Mathematics

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nirvana33 [79]1 year ago

5 0

17.8 you’re welcome. hyrewwtjiolkmbvxzawryjkkbcddtukpppiu

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Find the area of a regular decagon with an apothem of 5 meters and a side length of3.25 meter. Round to the nearest tenth

topjm [15]

We know that

The regular decagon can be divided into 10 congruent triangles
so
area regular decagon=10*[area of one triangle]

area triangle=b*h/2
where
b=3.25 m
h is the aphotem
h=5 m
area=3.25*5/2----> area=8.125 m²

area regular decagon=10*[8.125]----> 81.25 m²

Area=81.3 m²

the answer is
81.3 m²

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3 years ago

Carly and Simon are renting formal clothes for a wedding.

Cerrena [4.2K]

Answer:

C. y = 21x+90

Step-by-step explanation:

Let x be the number of hours. Each hour costs 21 dollars, so the cost of all the hours would be 21x. Then add a fixed fee of 90 to get C, y = 21x+90

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3 years ago

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Point Q is plotted on the coordinate grid. Point P is at (40, −20). Point R is vertically above point Q. It is at the same dista

lara31 [8.8K]

Answer:

$\displaystyle d_{RP}=50\sqrt{2}\ units$

Step-by-step explanation:

Distance between points in $R^2$

If P(p1,p2) and Q(q1,q2) are points on the plane $R^2$ , the distance between them is

$\displaystyle d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}$

We have Q(-10,-20) plotted on the coordinate grid. We also know that P is at (40, -20). We can see they have the same y-coordinate, so the distance between them is computed simply by subtracting their x-coordinates

$\displaystyle d_{PQ}=40-(-10)=50$

We must locate R knowing it's vertically above Q (x-coordinate = -10) and at the same distance from point Q as point P is from point Q. That means that from R to Q there are 50 units. They-coordinate of R will be -20+50=30.

The point R is located at (-10,30)

The distance from R to P is

$\displaystyle d_{RP}=\sqrt{(-10-40)^2+(30+20)^2}$