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3 years ago

Pic below................................ answer ASAP

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

7 0

Answer:

The answer is C.-216

Step-by-step explanation:

oee [108]3 years ago

7 0

Answer:

Step-by-step explanation:

Since -6 is negative, it makes it become a positive when multiplyed, and then back to a negative when multiplyed again.

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I need help with a problem with solving by square roots in quadratic equation.

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To solve for x, first, we add 3 to the given equation:

$\begin{gathered} 2(x+5)^2-3+3=44+3, \\ 2(x+5)^2=47. \end{gathered}$

Dividing by 2, we get:

$(x+5)^2=\frac{47}{2}.$

Therefore:

$x+5=\pm\sqrt{\frac{47}{2}}.$

Finally, subtracting 5 we get:

$x=\pm\sqrt{\frac{47}{2}}-5.$

Answer:

$x=\operatorname{\pm}\sqrt{\frac{47}{2}}-5.$

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1 year ago

Macy is painting a design that contains two repeating patterns. One pattern repeats every 8 inches. The other repeats every 12 i

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The pattern will be at the same place at 2 feet, 4 feet, 6 feet, 8 feet, 10 feet, 12 feet, 14 feet, 16 feet, and 18 feet

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3 years ago

Simon has a certain length of fencing to enclose a rectangular area. The function

TiliK225 [7]

Answer:

336

Step-by-step explanation:

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3 years ago

Joann can run the first 50 meters of the 200-meter race in 6.3 seconds. If she can maintain the same speed for the whole rece, h

ioda

Answer:

Step-by-step explanation:

shes fastr

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3 years ago

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).

4 0

3 years ago