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3 years ago

While performing the ″roll out the sunshine″ dance, what do your arms do at the very beginning of the sequence?

1 answer:

6 0

In performing the ″roll out the sunshine″ dance, we extend our arms wide. This mimics the radiance of the rays of the sun that extends infinitely in all the directions. The arms are then moving like a sun like a pulsating beat as sun rays do in animation.

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Assist with this please

jeka57 [31]

Answer: c should be your answer hope this helped

plz make brainly

Step-by-step explanation:

6 0

3 years ago

What is the area of the trapezoid to the nearest tenth? PICTURE BELOW ⬇⬇⬇⬇

Nastasia [14]

Answer:

62.4 ft²

Step-by-step explanation:

The unmarked horizontal dimension at the bottom of the triangle is ...

(8 ft)sin(30°) = 4 ft

The unmarked vertical dimension of the triangle (the height of the trapezoid) is ...

(8 ft)cos(30°) ≈ 6.93 ft

Then the area of the trapezoid is given by the formula ...

A = (1/2)(b1 +b2)h

A = (1/2)((4 ft+7 ft) +(7 ft))(6.93 ft) ≈ 62.4 ft²

_____

The mnemonic SOH CAH TOA can remind you of the relationships between right triangle dimensions and angles.

Sin = Opposite/Hypotenuse ⇒ Hypotenuse×Sin = Opposite

Cos = Adjacent/Hypotenuse ⇒ Hypotenuse×Cos = Adjacent

6 0

3 years ago

HURRY! WILL GIVE BRAINLIEST! HURRY

Helga [31]

 $answer \\ = - 0.5 \\ please \: see \: the \: attached \: picture \: for \: \\ full \: solution \\ hope \: it \: helps \\ good \: luck \: on \: your \: assignment$ 

5 0

3 years ago

Read 2 more answers

The distance between each base on a field is 90 ft. How far does a catcher have to throw the ball from home plate to second base

Gnoma [55]

Step-by-step explanation:

A baseball field is in the shape of a square (or diamond). Home plate and second base are opposite corners of each other. So the distance is equal to the diagonal of the square.

Use Pythagorean theorem to find the diagonal, or use properties of a 45-45-90 triangle.

Using Pythagorean theorem:

c² = a² + b²

c² = (90 ft)² + (90 ft)²

c = 90√2 ft

c ≈ 127 ft

7 0

4 years ago

Any 10th grader solve it 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

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

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

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

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

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

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