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Furkat [3]
3 years ago
10

Four-thirds (x) minus one-third = 9. Four-thirds (x) minus one-third + one-third = 9 + one-third. Four-thirds (x) = StartFractio

n 28 over 3 EndFraction
Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer:

x = 7

Step-by-step explanation:

Given

\frac{4}{3}x - \frac{1}{3} = 9

Required

Solve for x

\frac{4}{3}x - \frac{1}{3} = 9

Add \frac{1}{3} to both sides

\frac{4}{3}x - \frac{1}{3} + \frac{1}{3} = 9 + \frac{1}{3}

\frac{4}{3}x = 9 + \frac{1}{3}

Take LCM

\frac{4}{3}x = \frac{27 + 1}{3}

\frac{4}{3}x = \frac{28}{3}

Multiply both sides by \frac{3}{4}

\frac{3}{4} * \frac{4}{3}x = \frac{3}{4} *\frac{28}{3}

Evaluate the left hand side

x = \frac{3}{4} *\frac{28}{3}

x = \frac{28}{4}

Divide 28 by 4

x = 7

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Answer:

210 books

Step-by-step explanation:

Hello

I think  I can help you with this

Step 1

if one box is 7.5% of the total shipment, how many box is the total shipment

1 box =7.5%

x?box=100%

using a rule of three

\frac{1\ box}{7.5 } =\frac{x}{100}

Now, isolating x

\frac{1\ box*100}{7.5 } =x\\x=13.34\ boxes

13.34 boxes is the total

Step 2

find the total  of books

we will need 14 boxes, because it does not exist 0.34  box

so, the total of books=

number of boxes*number of books included in each box

=14 boxes*15 books

=210 books

Have a great day

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Read 2 more answers
How many equivalence relations are there on the set 1, 2, 3]?
Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0
3 years ago
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