All categories

3 years ago

Find the area of the regular decagon if the apothem is 2.8ft and a side is 1.8ft

1 answer:

6 0

DECA=10, decagon = 10 sides.

since each side is 1.8 long, all 10 sides will be 10*1.8 or 18 units long, so that'd be the perimeter then,

$\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{2}ap~~
\begin{cases}
a=apothem\\
p=perimeter\\
------\\
a=2.8\\
p=18
\end{cases}\implies A=\cfrac{1}{2}(2.8)(18)$

You might be interested in

Lizzie walked 8/10 mike and billy walked 12/15 mile. Lizzie says they walked the same distance.Do you agree with Lizzie? Explain

Alex_Xolod [135]

Answer:

true they walk the same distance

Step-by-step explanation:

8/10 is equal to 0.8

also 12/15 is also 0.8

therefore walking the same distance

7 0

3 years ago

Read 2 more answers

A different rectangle has a diagonal of 60 yards and a height of 40 yards.find the width of the rectangle rounded to the nearest

Debora [2.8K]

Answer:

2159.6 i think

Step-by-step explanation:

i used a rectangle calculator

5 0

2 years ago

Calculate work by parts , (-24)+(-8)

Triss [41]

(-32)

Step-by-step explanation:

(-24)+(-8)

= -24 -8

= -32

5 0

3 years ago

Read 2 more answers

Joel borrowed $1,512. He was to pay that amount back along with an additional 9% interest. He could only pay back 82% of what he

vodka [1.7K]

Answer:

$1351.43

Step-by-step explanation:

<u>The amount to pay:</u>

$1512 + 9% =
$1512*1.09 =
$1648.08

<u>The amount paid back:</u>

82% of $1648.08 =
0.82*1648.08 =
$1351.43

6 0

2 years ago

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Jlenok [28]

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

$r^2+2r+1=(r+1)^2=0\implies r=-1$

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$

4 0

3 years ago