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anastassius [24]
2 years ago
5

There are 1,570 souvenir paperweights that need to be packed in boxes. Each box will hold 17 paperweights. How many boxes will b

e needed? boxes will be needed to hold all the souvenir paperweights.
Mathematics
1 answer:
arlik [135]2 years ago
8 0

Answer:

93 boxes will be needed.

Step-by-step explanation:

Divide 1570 ÷ 17

This gives the answer 92.35

So 92 boxes won't be enough. 0.35 of a box itsn't helpful in the real world. So you need 93 boxes.

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If f(x) is discontinuous, determine the reason. f(x)= {x^2 +4, x<=1; x+3, x>1}
tester [92]

Answer:

Observe that f(x) is a continuous function when x1 because is a polynomial. The possible problem may occur in x=1.

Then, f(x) is discontinuous in x=1 if the limits of f to the right and the left of 1 exist and are different or if some of those limits doesn't exist.

Let's calculate the limits:

lim_{x\rightarrow 1^+}f(x)=lim_{x\rightarrow 1}(x+3)=1+3=4

lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^-}(x^2+4)=1^2+4=5

Since, lim_{x\rightarrow 1^-}\neq lim_{x\rightarrow 1^+} then f(x) is discontinuous in x=1.

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144 is a rational number
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2 years ago
Mason has a collection of 49¢ stamps, 20¢ stamps, and 34 stamps worth $23.55. He has 56 total 49¢ and 20+ stamps and the number
tatuchka [14]

Answer:

Mason has

  • 16 20¢ stamps;
  • 25 3¢ stamps;
  • 40 49¢ stamps.

Step-by-step explanation:

Let x be the number of 20¢ stamps, then the number of 3¢ stamps is  x + 9.

Mason has 56 total 49¢ and 20¢ stamps, then the number of 49¢ stamps is 56 - x.

Amounts:

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  • in 3¢: 3(x+9) cents.

Total amount:

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Hence,

49(56-x)+20x+3(x+9)=2,355\\ \\2,744-49x+20x+3x+27=2,355\\ \\-49x+3x+20x=2,355-2,744-27\\ \\-26x=-416\\ \\26x=416\\ \\x=16

Mason has

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3 years ago
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