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2 years ago

What 6x3 and 3x6 thankkkkkkkkkkkkkkkkk

Mathematics

2 answers:

Fed [463]2 years ago

8 0

Answer:

If you meant by (6x)³ and (3x)⁶.

(6x)³ = 6x · 6x · 6x = 108x³

(3x)⁶ = 3x · 3x · 3x · 3x · 3x · 3x = 729x⁶

If you meant by 6x³ and 3x⁶, then it is as it is.

If you meant 6 × 3 and 3 × 6,

6 × 3 = 18

3 × 6 = 18

6 × 3 = 3 × 6

yanalaym [24]2 years ago

5 0

Answer 6x3=18 and 3x6=18

Step-by-step explanation:

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Sally has just finished her thirty-fifth year with her company and is getting ready to retire. During her thirty-five years, Sal

olya-2409 [2.1K]

The <em>correct answer</em> is:

$19,153.26.

Explanation:

She will receive 42% of her average annual salary. This salary was $45,603.

To find 42% of a number, we first convert the percent to a decimal: 42% = 42/100 = 0.42.

Next we multiply this by the number we're taking the percent of:

0.42(45603) = 19153.26.

7 0

2 years ago

Read 2 more answers

Can y’all help me and tell me what the answer is please I’m a bit stuck

12345 [234]

Answer:

The correct choice is:

-2 to the 4th power is positive 16.

Option C is correct.

Step-by-step explanation:

We need to find mistake make in work below:

$(-2x^4)^4=-16x^{16}$

The mistake here is when we have even power, the negative sign gets positive i.e $(-a)^m=a^m$ if m is even

so, in our case it would be:

$(-2x^4)^4=16x^{16}$

The correct choice is:

-2 to the 4th power is positive 16.

Option C is correct.

5 0

2 years ago

A raffle offers one $8000.00 prize, one $4000.00 prize, and five $1600.00 prizes. There are 5000 tickets sold at $5 each. Find t

Harman [31]

Answer:

The expectation is $E(1 )= -\$ 1$

Step-by-step explanation:

From the question we are told that

The first offer is $x_1 = \$ 8000$

The second offer is $x_2 = \$ 4000$

The third offer is $\$ 1600$

The number of tickets is $n = 5000$

The price of each ticket is $p= \$ 5$

Generally expectation is mathematically represented as

$E(x)=\sum x * P(X = x )$

$P(X = x_1 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_1 ) = 0.0002$

Now

$P(X = x_2 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_2 ) = 0.0002$

Now

$P(X = x_3 ) = \frac{5}{5000}$ given that they offer five

$P(X = x_3 ) = 0.001$

Hence the expectation is evaluated as

$E(x)=8000 * 0.0002 + 4000 * 0.0002 + 1600 * 0.001$

$E(x)=\$ 4$

Now given that the price for a ticket is $\$ 5$

The actual expectation when price of ticket has been removed is

$E(1 )= 4- 5$

$E(1 )= -\$ 1$

4 0

2 years ago

112.324 nearest tenth

slavikrds [6]

Answer:

112.3

Step-by-step explanation:

112.3 if the 2 was 5 or higher it would be 112.4

5 0

3 years ago

Read 2 more answers

Can people help me with this

marishachu [46]

3) 21,070
5) 600
6) 265
I couldn't see #4
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6 0

3 years ago

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