The most which number has the same value as 20 tens

Question 1(Multiple Choice Worth 5 points)

Nadusha1986 [10]

Answer: $arc\ length =13.19cm$

Step-by-step explanation:

You need to use the following formula:

$arc\ length=r\theta$

Where "r" is the radius and $\theta$ is is the central angle of the arc in radians.

Assuming that that the angle $\frac{3\pi }{5}$ is the central angle of the arc, and knowing that the radius is 7 centimeters, you can substitute values into the formula.

Therefore, you get:

$arc\ length=r\theta\\\\arc\ length= (7cm)(\frac{3\pi}{5})}\\\\arc\ length =13.19cm$

4 0

3 years ago

1 tenth of 70kg is if you help thx

kkurt [141]

it would be 70/100 then it would go down to 7/10 so 1/10 of 70kg would be 7kg because you change 70 you make it a decimal witch is .70 or .7 now make that into a fraction and it would be 70/100 or 7/10 then you take your 7 because its 1/10 of 70. Or you could do it a whole lot easier way and just divide 70 by 10 and get 7. There you have it your answer is 7.

HOPE THIS HELPED!!!

THANKS!!!

8 0

3 years ago

Am I right?

Yuliya22 [10]

I think this MIGHT be right I’m not sure but yeah

5 0

3 years ago

Read 2 more answers

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

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

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

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

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

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

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