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4 years ago

A scientist is working with 19 meters of gold wire. How long is the wire in millimeters?

Mathematics

2 answers:

Elza [17]4 years ago

8 0

Answer

19000 mm

Step-by-step explanation:

To convert meters to millimeters

Multiply the number by 1000 to get your answer

In this case.

19mx1000= 19000mm

Vadim26 [7]4 years ago

7 0

Answer: what they said

Step-by-step explanation:

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the ez pay plan costs $0.15 per minute. write an expression that represents the monthly bill for x minutes on the ez pay plan

Neporo4naja [7]

we are given

The ez pay plan costs $0.15 per minute

so, cost is $0.15 per minutes

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x minutes on the ez pay plan

so, time for plan is x minutes

total bill = cost * time for plan

now, we can plug values

we get

total bill = $0.15x..............Answer

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3 years ago

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Someone help me PRETTY PLEASE :(((!!!!!!!!!!

Alecsey [184]

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3 years ago

Please need answer help help?!?!?

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Answer:

Step-by-step explanation:

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3 years ago

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

4 years ago

Please solve with explanation

Juliette [100K]

Step-by-step explanation:

$4\frac{9}{20} - 1\frac{2}{6} =\\\frac{89}{20} - \frac{8}{6} = \frac{89 - 26.7}{20} = \frac{62.3}{20} \\$

$4\frac{6}{9} - 3\frac{3}{6} =\\\frac{42}{9} - \frac{21}{6} = \frac{252 - 189}{54} = \frac{63}{54} = 1\frac{9}{54}$

$2\frac{2}{4} - 1\frac{1}{3} =\\\frac{10}{4} - \frac{4}{3} = \frac{30 - 16}{12} = \frac{14}{12} = 1\frac{2}{12}$

8 0

4 years ago

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