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2 years ago

What is the Kinetic Energy of a 15 kg object that is moving with a speed of 30 m/s?

Mathematics

2 answers:

Eddi Din [679]2 years ago

7 0

Answer:

K=6750[J]

Step-by-step explanation:

$Kinetic \:Energy=\frac{1}{2} mv^2\\\\K=\frac{1}{2} (15)(30)^2\\\\K=6750[J]$

crimeas [40]2 years ago

5 0

Answer:

6750J

Step-by-step explanation:

30^2=900 900x15=13500 13500/2=6750

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What is the inverse of y=-x-4

Nimfa-mama [501]

Answer:

$f^-1 -1(x)=x+4$

Step-by-step explanation:

i hope this helps

8 0

2 years ago

Need help ASAP!<br> Which inequality is represented by this graph

Nadusha1986 [10]

Answer:

the first answer

Step-by-step explanation:

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3 years ago

6. Steve is packing sacks for treats at a party. Every sack is exactly the same. Steve has 42 candy bars and 126 lollipops. a. W

Sergeu [11.5K]

Answer:

Greatest sack = 42

1 candy bar and 3 lollipops

Step-by-step explanation:

Represent Candy bars with C and Lollipops with L

$C = 42$

$L= 126$

Solving (a): Greatest number of treat sacks

To solve this, we simply calculate the GCF of C and L

$42 = 2^1 * 3^1 * 7^1$

$126 = 2^1 * 3^2 * 7^1$

Hence, the GCF is

$GCF = 2^1 * 3^1 * 7^1$

$GCF = 42$

Hence, greatest number of sack is 42

Solving (b): Number of treat in each sack.

To do this, we simply divide the number of C and L by the calculated GCF

For C:

$Treats = \frac{C}{GCF}$

$Treats = \frac{42}{42}$

$Treats = 1$

For L:

$Treats = \frac{L}{GCF}$

$Treats = \frac{126}{42}$

$Treats = 3$

<em>Hence, 1 candy bar and 3 lollipops</em>

8 0

2 years ago

THIS IS URGENT!<br> add 7-9=7+(-9)

Burka [1]

Answer:

-3=3

<h2><em>Hope this helps!!!</em></h2>

7 0

3 years ago

Read 2 more answers

Students are doing an experiment: that two fair dice are tossed, and all the outcomes are recorded.Then,an event is described as

Marysya12 [62]

Answer:

1.The random variable X can take any value from( 2,- ----,12) in the event of the sample space.

2.The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3), (2, 2),(3, 1), (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6), (2, 5),(3, 4),(4, 3),(5, 2), (6, 1), (2, 6),(3, 5), (4, 4), (5, 3), (6, 2),(3, 6), (4, 5), (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6), (6,5),(6,6)]

3. The probability of getting a 4 = 1/12

4. The probability of getting a 12 = 1/36

Step-by-step explanation:

There are 36 outcomes in all and the event shows the possible numbers obtained after the sum .

The event will be E= [ 2,3,4,5,6,7,8,9,10,11,12]

The random variable X can take any value from( 2,- ----,12) in the event of the sample space.

The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3), (2, 2),(3, 1), (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6), (2, 5),(3, 4),(4, 3),(5, 2), (6, 1), (2, 6),(3, 5), (4, 4), (5, 3), (6, 2),(3, 6), (4, 5), (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6), (6,5),(6,6)]

that is it has all the possible outcomes of both the dice.

The probability of getting a 4= Number of fav outcome/ no of possible outcomes= 3/ 36= 1/12

The probability of getting a 12= Number of fav outcome/ no of possible outcomes=1/36

6 0

2 years ago