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DerKrebs [107]
3 years ago
13

3. The figure below is a composite shape that consists of a semicircle and an equilateral triangle (triangle

Mathematics
1 answer:
Ivenika [448]3 years ago
8 0

Answer:

See explanation below

Step-by-step explanation:

Area = Area of triangle + Area of the semi circle

Area of triangle = 1/2bh

Area of triangle = 1/2 *(5/2)(4)

Area of triangle = 1/2(2.5)*4

Area of triangle = 2 * 2.5

Area of triangle = 5cm^2

Area of semicircle = πr²/2

r = 2.5cm

Area of semicircle =(3.14)(2.5)²/2

Area of semicircle = = 3.14 * 4.5

Area of semicircle = 9.8125cm^2

Area pf the figure = 5+9.8125

Area pf the figure = 14.8125cm^2

For the perimeter

Perimeter of the semicircle = 2πr

Perimeter of the semicircle = 2(3.14)(2.5)

Perimeter of the semicircle = 15.7cm

Perimeter of the semicircle = = 15.7 + 5 + 5

Perimeter of the semicircle = 25.7cm

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Any 10th grader solve it <br>for 50 points​
kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., a_{1}=A and common difference=D

The nth term can be written as

a_{n}=A+(n-1)D

pth term of given arithmetic progression is a

a_{p}=A+(p-1)D=a

qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

rth term of given arithmetic progression is c

a_{r}=A+(r-1)D=c

We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
3 years ago
Working alone, Darryl can mop a warehouse in 11 hours. Elisa can mop the same warehouse in 9
Harrizon [31]

Answer:

  • ≈ 5 hours

Step-by-step explanation:

<u>Time needed if they work together:</u>

  • 1 / (1/11 + 1/9) =
  • 1/ (20 / 99) =
  • 99/20 ≈ 5 hours
3 0
2 years ago
PLEASE HELP ME!!!
crimeas [40]
B=4676+10043 c=b-2160 answer equals b+c
I hope this is helpful I am trying not to say the exact answer but to teach you instead
4 0
3 years ago
Owen is signing up for a gym membership with a one-time fee to join and then a
Katena32 [7]
The cost of the monthly fee is 50$.. if the joining fee is 100.. take 200 and subtract that fee since it is included in the total. You’re left with 100. Note, the 200 is calculated for a 2 month period, so split the remaining 100 ,from the total 200, into 2. You’re left with 50 each month.

T=(C-100)/2 (T being 1 month) (C being the total cost/200)

Hopefully

3 0
3 years ago
In which number is the value represented by the digit 9 equal to 110
Tresset [83]

Answer:

49

Step-by-step explanation:

a+b/2xh

aksfsaowcapa

8 0
3 years ago
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