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3 years ago

What is 50/7 as a mixted number

Mathematics

2 answers:

sattari [20]3 years ago

8 0

Answer: 7 1/7

Step-by-step explanation:

Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.

soldier1979 [14.2K]3 years ago

6 0

Answer:

7 $\frac{1}{7}$

Step-by-step explanation:

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Todd is saving for a vacation. The cost of his vacation cost $1,089. Todd has a year to save the money. About how much does he n

Serggg [28]

Answer:

He needs to save 90.75

Step-by-step explanation:

Take the money the vacation will cost and divide it by the number of months in a year. 1,089 ÷ 12= 90.75

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3 years ago

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PLEASE HELP ME IM TAKING TIMED TEST I WILL MARK YOU AS BRAINLEST!!

stepladder [879]

Answer:

-4.5,-0.5,0.3,0.7,2.3

Step-by-step expla

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3 years ago

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Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

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3 years ago

Jaqueline used 2.5 pounds of ground beef to make 25 tacos for a family gathering. Peter wants to use the same recipe using 1 pou

enot [183]

Answer:

Peter can make 10 tacos.

Step-by-step explanation:

Jaqueline's recipe calls for .1 pounds of beef per taco.

Given only 1 pound, multiply by, taking the reciprocal of .1 gives us 10 tacos.

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3 years ago

Simplify (-5) + (-4) +(7)

MAXImum [283]

Answer:

Lol, isn't that just -2?

Step-by-step explanation:

I think you can just add those up.

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3 years ago

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