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Levart [38]
2 years ago
7

A game decreased in price by 2/3. After the reduction it was priced at £13. What was the original price of the game?​

Mathematics
2 answers:
Nezavi [6.7K]2 years ago
7 0

Answer:

£19.50

Step-by-step explanation:

let x be the original price , then

\frac{2}{3} x = 13 ( multiply both sides by 3 to clear the fraction )

2x = 39 ( divide both sides by 2 )

x = 19.5

original price of game was £19.50

Archy [21]2 years ago
5 0

Answer:

x = 39

Step-by-step explanation:

Based on the given conditions, write: x=\frac{13}{1-\frac{2}{3}}

Find common denominator and write numerators above common denominator: x=\frac{\frac{13}{3-2} }{3}

Calculate the sum or difference: x=\frac{\frac{13}{1} }{3}

Divide the fraction by multiplying its reciprocal: x=13*3

Calculate the product or quotient: x=39

Answer: x=39

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An unknown fraction is between 0 and 1. Which is true about the equivalent percent and decimal?
pashok25 [27]

Answer:

The answer is C.

Step-by-step explanation:

The number between 0 and 1. Think of decimals as cents, 100 cents makes a whole. So the number will be greater than 0 but less than 100.

7 0
3 years ago
John has all the marbles in the world. What percent of the worlds marbles does John have?
Eddi Din [679]

Answer:

100%

Step-by-step explanation:

All of them = 100%.

7 0
3 years ago
Can some one do this and help me out
alexira [117]

Answer:

<h2><em><u>4</u></em></h2>

Step-by-step explanation:

Numbers less than 10 are = 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Not multiple of 2 = 3, 5, 7, and 9

Composite number = 4

I eliminated the numbers to get the answer.

Hope this helped!

5 0
3 years ago
1/4 the difference of 4/6 + 3/12
Soloha48 [4]
The answer would be about 8/12 because you must convert 4/6 into 8/12. 8/12 + 3/12=11/12 
1/4 would be about  3/12, so 11/12 - 3/12=8/12.
5 0
3 years ago
Suppose you can factor x^2 +bx +c as (x+p)(x+q) . If c&lt;0, what could be the possible values of p and q?
Anna11 [10]
Ans: Option A

Explanation:
Let's solve it smartly!
Given expression: x^{2} + bx +c --- (A)
Factors: (x+p)(x+q)
Condition: c<0
Now let us expand (x+p)(x+q):
=> x^{2} + (p+q)x + pq  --- (B)

By comparing (B) with (A), we can say that:
pq = c --- (C)

Now, as the condition says, c<0, it means either p or q is negative. Both cannot be positive or both cannot be negative.

1) If p>0, q>0, it means c>0 since (+p)(+q) = (+c)(according to equation (C)). Condition is not met.
Hence, option B and D are wrong.

2) If p<0, q<0 it means c>=0 since (-p)(-q) = (+c)(according to equation (C)). Condition is not met.
Hence option C is out as well.

We are left with Option A:p<0, q>0 it means c<0 since (-p)(+q) = (-c)(according to equation (C)).
Condition is MET!
Hence,
Ans: Option A: p= -3, q= 7
4 0
3 years ago
Read 2 more answers
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