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3 years ago

What is a equivalent fraction for 6/8

Mathematics

2 answers:

lesya [120]3 years ago

7 0

Answer:

3 / 4

Explanation:

Find a number that divides evenly into both 6 & 8

Number: 2

6 ÷ 2 = 3

8 ÷ 2 = 4

= 3 / 4

Paladinen [302]3 years ago

5 0

Answer:

3/4, 12/16

Step-by-step explanation:

times two or divide two

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Grace [21]

Answer:

a true

b false

c. false

d. true

e. false

f. true

Step-by-step explanation:

The function is y =35+8x where x is the number of items sold and y is the earnings

To determine if the point is on the graph substitute the point in and see if it is true

a (7,91)

91 = 35+ 8(7)

91 = 35+ 56

91 = 91 true

b(11, 121)

121 = 35+ 8(11)

121 = 35+ 88

121 = 123 false

c. (9, 110)

110 = 35+ 8(9)

110 = 35+ 72

110 = 107 false

d. (4,67)

67 = 35+ 8(4)

67 = 35+ 32

67 = 67 true

e. (5,72)

72 = 35+ 8(5)

72 = 35+ 40

72 = 75 false

f. (3, 59)

59 = 35+ 8(3)

59 = 35+ 24

59 = 59 true

8 0

3 years ago

Please help me please

n200080 [17]

Answer:

Step-by-step explanation:

110_5

7 0

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

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

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

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

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

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

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