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Anettt [7]
3 years ago
13

Mr.jone wants to buy 6 tickets to a play.Each ticket costs $10 Mr.Jone has $45 in his pocket. How much more money does he need t

o buy 6 tickets?
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
8 0
He needs $15 more to buy the tickets.
ohaa [14]3 years ago
5 0
6 tickets at $10 a piece is $60 total.
$60 - $45 = $15
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Solve the systems by elimination: <br><br> 8x-2y= -10<br> -4x+y= 5
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Let's eliminate x.  To achieve this, mult. the 2nd equation by 2, obtaining

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4 0
3 years ago
Julie says I added three odd numbers and my answer was 50 explain why Julie cannot be correct
ANTONII [103]
Julie cannot be right. That's because if you add three odd numbers together, <em>you always get an odd number. </em>The number 50 is an Even number.

Let's try adding 3 odd numbers together a couple of times...
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8 0
3 years ago
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4. Find the H.C.F. of the following numbers. <br>6. 68,48 b. 64, 128<br>c. 24,54​
Goryan [66]

Answer:

Step-by-step explanation:

a)Factors of 6 = 1, 2, 3 and 6. Factors of 9 = 1, 3 and 9. Therefore, common factor of 6 and 9 = 1 and 3. Highest common factor (H.C.F) of 6 and 9 = 3.

he HCF of 68 is 68.....! bcoz 68 itself is a highest factor of 68.....!

48 are 2×2×2×2⇒16

b)64 = 1 × 2 × 2 × 2 × 2 × 2 × 2. 80 = 1 × 2 × 2 × 2 × 2 × 5.

Factors of 128 are 1, 2, 4, 8, 16, 32, 64, 128. There are 8 integers that are factors of 128. The greatest factor of 128 is 128. 3.

c.Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

54 = 2 × 27 = 2 × 3 × 9 = 2 × 3 × 3 × 3. 81 = 3 × 27 = 3 × 3× 9 = 3 × 3 × 3 × 3. 99 = 11 × 9 = 11 × 3 × 3. Factors common to all three numbers is 3 × 3.

8 0
3 years ago
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