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3 years ago

My sandbox is a square each side is 6 feet long what is the area of the sandbox

Mathematics

1 answer:

alexgriva [62]3 years ago

8 0

Answer:

36 feet²

Step-by-step explanation:

Each side is 6 feet

Multiply length * width to find area

6 * 6

36 feet²

Hope this helps :)

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How to tell if a triangle is obtuse acute or right based on side lengths?

tekilochka [14]

If a triangle is made up of side lengths a, b and c

Given that c is the longest side length of the triangle,

The Pythagorean theorem can be used to tell if the triangle is obtuse, acute or right

In an obtuse triangle,

a^2 + b^2 < c^2

In an acute triangle,

a^2 + b^2 > c^2

In a right angled triangle,

a^2 + b^2 = c^2

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3 years ago

Equivalent fractions of 2 6th

marta [7]

2/6 is also equal to 1/3. This is because the GCF of 2 and 6 is 2, and dividing both the numerator and denominator by 2 gets us a simplified fraction of 1/3.

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4 years ago

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bulgar [2K]

Answer: he had 5 pounds.

Step-by-step explanation:

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3 years ago

N=r (A-s) what is s

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3 years ago

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

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

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

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

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

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

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