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3 years ago

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If your weight is 150 pounds and your mass is 15.3 kilograms, what would those values be if you were on the moon? The gravitatio

nal force on the moon is 1/6 the gravitational force on Earth.

1 answer:

natulia [17]3 years ago

3 0

Weight = mass * g where g = gravitational force

so a weight of 150 pounds would be 150 / 6 = 25 on the moon.
The mass would be the same as on earth.

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X=17-4y<br> y=x-2<br> answer by substitution

sashaice [31]

Answer:

x=5,y=3

Step-by-step explanation:

the first step is to decide whether you want to substitute y and find x, or substitute x and find y.

(it doesn't matter which way you will choose.If you do the math correctly you will get the correct answer,one way or another).

for example, let's substitute y into the first equation:

x=17-4*(x-2)

so now,we will organize the equation:

x+4x=17+8

5x=25/(÷5)

x=5

now we will substitute x with 5 in the second equation(or the first one,it doesn't matter)in order to find y:

y=5-2

y=3

3 0

3 years ago

A code is formed using four of these letters: A, B, I, K, N, O, and T. (Note that the order of the letters in the code matters;

oksano4ka [1.4K]

To answer this question, use permutation since the order of the letter matters. There are 7 letters and only 4 are to be used for the code. This is a permutation of 7 taken 4 (7P4) which is equal to 840. Thus, the sample space is 840.
For the next questions, there are 3 vowels for as the first letter, only 6 letters are available for the second, 5 on the next, and 4 on the last. Multiplying these numbers give 360. To solve for probability, divide this number with the sample space. Thus, the answer is 3/7.

7 0

3 years ago

Which table does NOT represent a linear function?

chubhunter [2.5K]

Hey there!
Linear functions have a continuous change.

Let's check these tables and see if we can tell linear functions from non-linear functions.

The first one is

x values: -1, 0, 1, 2

- we add 1 each time

y values: 8, 5, 2, -1

- we subtract 3 each time

$\rule{200}{2}$

Let's try the next one:

x values: -1, 0, 1, 2

- we add 1 each time

y values: 5, 10, 15, 20

- we add 5 each time

$\rule{200}{2}$

Let's try the third one:

x values: -1, 0, 1, 2
- we add 1 each time

y values: 15, 18, 20, 21

- we add 3, then 2, then 1..

So this table doesn't represent a linear function.

Let's check the fourth one:

x values: -1, 0, 1, 2

- we add 1 each time

y values: 1, 2, 3, 4

- we add 1 each time

Thus, Option C is the right option.

Hope everything is clear.

Let me know if you have any questions!

Always remember: Knowledge is power!

4 0

2 years ago

A bandleader randomly select's two students out of 100 bandmembers if you and your brother are both bandmembers what is the prob

kherson [118]

<span>There are 100 * 99 = 9900 different ways to choose two students out of 100 bandmembers. You and our brother are 2 of that ways: you being selected first and you being selected second. Therefore the probability that you two are selected is 2 / 9900 = 0,000202. You can also think as the probability of you being selected among 100 bandmembers, which is 1/100, times the probability of your brother being selected among 99 members, which is 1/99 => (1/100) * (1/99) = 1/ (100*99) = 1 / 9900; plus the same for your brother being selected first and you second => [1/9900] * 2 = 2/9900, which is the same calculated above.</span>

6 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago