Please help problem is in the image.

The additional cost per month of a sunroof, spoiler and stereo is represented by three consecutive even integers, respectfully.

lesya [120]

Answer:

16

Step-by-step explanation:

First of all, I think it was "respectively", not "respecfully", =))). A noticeable point.

From the problem, three item's costs are represented by three consecutive even integer.

Thus, we can call cost of spoiler is x, sunroof is x-2, stereo is x + 2. It is conditioned that x is an even number.

Since the sum of the three is 48, we have

(x-2) + x + (x+2) = 48

x+x+x -2 +2 = 48

3x =48

x=16

So the spoiler is 16; thus, the sunroof is 14 and the stereo is 18

3 0

3 years ago

The length of the minute hand is 150% of the length of the hour hand. In 2 hours, how much farther does the tip of the minute ha

nekit [7.7K]

The question seems incomplete, as the length of the hour hand isn't given ;

If the tip of the hour hand is taken as 30mm

Answer:

534.1 mm

Step-by-step explanation:

In 2 hours ; cycle performed :

2/ 12

Circumference of a circle = 2 * pi *r

Circumference = 2 * pi * 30 = 60pi

Distance traveled : 60pi * 2/12 = 120pi/ 12 =10pi

The minute hand travels 2 comply cycle in 2 hours

150% taller than short hand :

1.5 * 30 = 45 mm

Distance traveled : 2 (2 * pi * 45) = 180pi

(180pi - 10pi) = 170pi

170π = 534.07075

= 534.1 mm

7 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

An angle measures 10° less than the measure of its complementary angle. What is the measure of each angle?

Wewaii [24]

Answer:

50º & 40º

Step-by-step explanation:

complementary anlges add up to 90º. so, we can make an equation:

x+x-10=90

2x=100

x=50

so the angles are 50º and 40º.

5 0

3 years ago

. Simplify: (2 x 2 + 6-10 ÷<br><br> 2)

Maurinko [17]

Answer: 5

Step-by-step explanation:

You see 2x2 = 4 right and since this is operation 2 / 10 = 5 and 6-5 = 1 so add 1 and that gives you 5

2x2 = 4 / 10/2 = 5 / 6-5 = 1 +4 = 5

7 0

2 years ago

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