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3 years ago

6 min to 1 hour as a fraction in simplest form

Mathematics

2 answers:

andreev551 [17]3 years ago

4 0

Answer:1/10

Step-by-step explanation:

Alika [10]3 years ago

3 0

Step-by-step explanation:

$1 \: hour \: = 60 \: min \\ \\ \frac{6}{60} = \frac{1}{10} \\$

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Sara drove 96 miles on 3.2 gallons of gas. How many miles can Sara drive on 12.5 gallons of gas

Blababa [14]

Answer:

375 miles

Step-by-step explanation:

Create a proportion where x is the number of miles she can drive on 12.5 gallons of gas:

$\frac{96}{3.2}$ = $\frac{x}{12.5}$

Cross multiply and solve for x:

3.2x = 1200

x = 375

So, with 12.5 gallons of gas, Sara can drive 375 miles

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3 years ago

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An acid is prepared by mixing 400 mL of water and 100 mL of pure sulfuric acid. What is the

Natasha2012 [34]

C. 25
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Divide 400/100=4
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3 years ago

Factorise 2x²-9x+10 (this must be put in two brackets)

galben [10]

Find both roots :
$\Delta=81-80=1$ hence $r=\frac{9\pm1}4=\frac{5}2\text{ or }2$

Thus it is $2(x-\frac{5}2)(x-2)$

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3 years ago

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F y = 16x, which table shows the values of y for different values of x?

Natali5045456 [20]

X )0 2 4 y)0 16 32 is ur answer

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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

3 years ago