All categories

4 years ago

Work out ( 2 × 10 5 ) × ( 3 × 10 4 ) Give your answer in standard form.

Mathematics

2 answers:

AnnyKZ [126]4 years ago

4 0

Answer:

6 × 10^9

Step-by-step explanation:

( 2 × 10^5 ) × ( 3 × 10^4 )

Rearrange.

2 × 3 × 10^5 × 10^4

Multiply and apply the law of exponents.

6 × 10^(5+4)

6 × 10^9

frutty [35]4 years ago

4 0

Hello, I'd like to answer your question. So first, you write an equation for this question:

Equation: (2×10^5)×(3×10^4)

(2×10^5)×(3×10^4), then here are the steps:

Always use the Order of Operation.

() - Parenthesis

^x - Exponents

×/÷ - Multiplication/Division

+/- - Addition/Subtraction

So, in the parenthesis, there is an exponent. We solve that first.

10^5=100,000

10^4=10,000

Now, multiply these two.

(2×100,000)=200,000

(3x10,000)=30,000

Now, the answer is:

200,000×30,000=6000000000

Thus, we can write it simplified.

6×10^9 is the answer.

You might be interested in

If RT is greater than BA, which correctly compares c and s

Fofino [41]

Answer:

Step-by-step explanation:

The answer is B

By the converse of the hinge theorem, mC = mS. FALSE. EITHER ANGLE IS LARGER THAN THE OTHER

By the hinge theorem,TS > AC. FALSE. CONGRUENT SIDES BASED ON THE GIVEN IMAGE.

By the converse of the hinge theorem, mS > mC.the statement that is true

By the hinge theorem, BA = RT. FALSE. 3RD SIDE OF EITHER TRIANGLE WILL BE LONGER THAN THE OTHER.

3 0

3 years ago

Which expression is equivalent to -6(-2/3+2x) <br> -4-12x<br> -4+2x <br> 4-12x<br> 4+12x

Basile [38]

C. Is your answer to the question

4 0

3 years ago

Read 2 more answers

All rational numbers are also? A) real numbers b)natural numbers c)integers d)whole number

Alex Ar [27]

Pretty sure it's C)intergers

8 0

3 years ago

Read 2 more answers

Kenidi is shopping for a couch with her dad and hears him ask the salesperson, "How much is your commission?"The salesperson say

denis23 [38]

Answer:37.15

Step-by-step explanation:

6 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