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pantera1 [17]
3 years ago
6

Pls help guys I’m desperate and I need to go to bed

Mathematics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

R has to be 8 or above

Step-by-step explanation:

Hope this helps :)

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Special precision bolts are made by another machine. These are for jobs that require more exact measurements. The bolts must be
klio [65]

The solutions of the inequality is 7.99<x<8.01. I think this solutions means the radius of hole in the bolt or something along that, but I'm not sure.

3 0
3 years ago
Solve for g : 3+2g=5g-9
KiRa [710]
3+2g=5g-9
5g-2g=3+9
3g=12
3g/3=12/3
g=4
5 0
3 years ago
8.2<br> 5. In a '70% off' sale, a hat was £33.60.<br> Work out the original price.
Aleks [24]

100% - 70% = 30%

30% of what = £33.60?

Let p = original price

0.30p = 33.60

p = 33.60 ÷ 0.30

p = £112

3 0
3 years ago
En una tienda de ropa vas a comprarte una pantalones que están de 150 a 90 euros y una camiseta de 40 a 24 ¿ tienen el mismo por
uranmaximum [27]

Respuesta:

Sí, tienen el mismo descuento del 40%

Explicación paso a paso:

Dado :

PANTALÓN :

Precio inicial = 150

Precio con descuento = 90

Porcentaje de descuento en el precio:

(150 - 90) / 150 * 100%

60/150 * 100%

0,4 * 100

= 40%

CAMISA:

Precio inicial = 40

Precio con descuento = 24

Porcentaje de descuento en el precio:

(40-24) / 150 * 100%

16/40 * 100%

0,4 * 100

= 40%

Sí, tienen el mismo descuento del 40%

8 0
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
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