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4 years ago

If y varies directly with x and y = 6 and x = -3, find y when x = 4.

Mathematics

2 answers:

Ber [7]4 years ago

8 0

Ask questions in the comment section if you have questions. I'm happy to help! <3

Step-by-step explanation:

What the question is telling you is that x and y and proportional, almost like a ratio.

Now we have to find y when x is four.

Let's create a proportional equation...

What can we divide by four to create -1/2?

Citrus2011 [14]4 years ago

6 0

Answer:

y = -8

Step-by-step explanation:

6/-3 = -2

4 * -2 = -8

y = -8

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When the denominator and numerator are equal, it means?

WITCHER [35]

The fraction is one whole aka one

5 0

3 years ago

The toy shop stocks tricycles and go-carts.

Mandarinka [93]

Total number of wheels in the toy shop = 37 wheels

Number of wheels for tricycles = 3

Number of wheels for go-carts = 5

Since the total number of tricycles and go-carts are not given, we solve this question using Assumption.

Let's assume: We have 4 tricycles and 4 go-carts

(4 x 3) + (4 x 5)

= 12 + 20

= 32 wheels. This assumption is wrong.

Let's assume: We have 4 tricycles and 5 go-carts

(4 x 3) + (5 x 5)

= 12 + 25

= 37 wheels. This assumption is correct.

Therefore, we have 4 tricycles and 5 go-carts in the toy shop.

To learn more, visit the link below:

brainly.com/question/23264573

3 0

3 years ago

How do you solve this equation thankyou. <br> 5n+34=-2(1-7n)

JulijaS [17]

$5n+34=-2(1-7n) \\
5n+34=-2+14n\\
-9n=-36\\
n=4$

5 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

Write each number in standard notation.

Zarrin [17]

Answer:

See Bottom!

Step-by-step explanation:

17) 9

18) 2.0000

19) 20000000000

20) 00.804

21) 266.0000000000

22) 15.00

23) 77.5

24) 830.0000000000

5 0

3 years ago