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2 years ago

When x =6 y=2 What is: 2(3x-6y)^2

Mathematics

2 answers:

sashaice [31]2 years ago

6 0

Answer:

$2(3x - 6y) ^{2} = 72$

Step-by-step explanation:

$2(3x - 6y) ^{2}$

$x = 6 \: \: y = 2$

Substitute in the values

$2(3(6) - 6(2))^{2}$

$2(18 - 12) ^{2}$

$2(6) ^{2}$

$2(36)$

$= 72$

ikadub [295]2 years ago

3 0

Answer:

Step-by-step explanation:

We first substitute x = 6 and y = 2 into the equation

2(3 x (6) - 6 x (2)) ^ 2

First we multiply all the numbers in the bracket

2(18 - 12) ^ 2

Now we minus the numbers in the bracket

2(6)^2

Now we use the indice rule to do 6^2 which is 36

And then we multiply 36 by 2 which is 72

Hope it helps :)

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A large box of jelly beans weighs 96.58 lb the jelly beans are evenly divided into 22 bags what is the weight in pounds of the j

Vadim26 [7]

Answer:

The weight in each bag is 4.39lb.

Step-by-step explanation:

To solve this, you simply need to divide 96.58 by 22. The reason you do this is so that way you can divide all of the weight into 22 bags.

96.58 ÷ 22 = 4.39

This means that the weight in each bag is 4.39lb.

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2 years ago

If Ann travels 50 meters North, 200 meters East and 150 meters South, what is her total distance travelled ? A. 150 meters B.400

olga2289 [7]

Answer:

400 meters

Step-by-step explanation:

50 + 200 + 150 = 400

3 0

3 years ago

Read 2 more answers

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

What is the result of adding

Aliun [14]

Answer: You get an answer?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

It has a slope of -7 and passestrough the point (-3,16)

Jlenok [28]

Answer:

y = -7x - 5

Step-by-step explanation:

y-intercept= 16 - (-7) (-3) = - 5

8 0

3 years ago