The potential energy, P, in a spring is represented using the formula P =1/2 kx2. Lupe uses an equivalent equation, which is solved for k, to determine the answers to her homework.Which equation should she use?
Answers
Answer 1
Answer:
The answer to this question will be K=2PX^2
Answer 2
Answer:
Answer:
the ∧ is a 2 square
Step-by-step explanation:
Related Questions
20 is 12.5$ of what number?
Question 1(04.03 LC)Which of the following options is an equivalent function to f(x) = 2(5)2X? (2 points)a. f(x) = 50xb. f(x) = 100xc. f(x) = 2(25)d. f(x) = 4(25)
Solve for x: 4 = (-8) +3x. Will you please help?
42) According to the 2000 Census the population of Canton was 7,720. Holly Spring's population was 3,200. Canton's population increases at a rate of 80 people a year and Holly Spring's population increases by 120 people a year. Which equation may be used to determine the number of years before the populations are equal?
Evaluate the exact value: tan(pi/6)
Question 1(04.03 LC)Which of the following options is an equivalent function to f(x) = 2(5)2X? (2 points)a. f(x) = 50xb. f(x) = 100xc. f(x) = 2(25)d. f(x) = 4(25)
Solve for x: 4 = (-8) +3x. Will you please help?
42) According to the 2000 Census the population of Canton was 7,720. Holly Spring's population was 3,200. Canton's population increases at a rate of 80 people a year and Holly Spring's population increases by 120 people a year. Which equation may be used to determine the number of years before the populations are equal?
Evaluate the exact value: tan(pi/6)
Answers
ANSWER: 98.90
hope that helps
hope that helps
Answer:
It's already there or it's not then put in 100
Step-by-step explanation:
Answers
There is a certain way to find the answer for your question. First, you must be familiar with the divisibility rule for 11: the difference between the sums of even digits and odd digits should either be 0 or 11 only. For example, 1430, here the odd digits are 4 and 0 (which are the 3rd and 1st digits); the even digits on the other hand are 1 and 3, placed as the 4th and 2nd digits respectively.
Adding the sums of odd and even digits will give 4 (4+0) and 4 (3+1). Now, the difference of these sums is obviously equal to 0. Therefore, 1430 is divisible by 11. Same case will hold for numbers whose sums of odd and even digits is 11.
Moving on, now let us try to find the biggest 6-digit number divisible by 11. Intuitively, the last digit should be "9" - the hundred thousands digit - for it is the biggest number among numbers from 0-9. Furthermore, since there is no restriction as to whether the same number could or could not repeat, we can still use "9" all throughout the remaining 5 digits (ten thousands down to ones digit), giving the greatest possible 6-digit number, 999,999.
To check if the number we have obtained, 999,999, is divisible by 11, we need to use the divisibility rule for 11. Sum of even digits is 18 (9+9+9); sum of odd digits is 18 (9+9+9) as well. The difference of the sums is 0. Therefore, 999,999 is divisible by 11, hence the greatest 6-digit number multiple of 11.
As for the 6-digit number whose digits' sum is 40 and multiple of 11, we can assume its first 4 digits to be as 999,9** - four 9s is already equal to 36 - we only need to find two numbers whose sum is 4 to make the sum's total be 40. These remaining numbers should also be divisible by 11 since 9999 is already divisible by 11. The remaining numbers should only be 2 and 2 since the sum of its digit is 4, and that if we put them together, 22, a multiple of 11. This is the only combination of numbers that will give a sum of 4 while maintaining its structure a multiple of 11.
At last, the greatest 6-digit number whose digits' sum equal to 40 and is a multiple of 11 is 999,922. True, for 999,922 / 11 is equal to 90,902.
Adding the sums of odd and even digits will give 4 (4+0) and 4 (3+1). Now, the difference of these sums is obviously equal to 0. Therefore, 1430 is divisible by 11. Same case will hold for numbers whose sums of odd and even digits is 11.
Moving on, now let us try to find the biggest 6-digit number divisible by 11. Intuitively, the last digit should be "9" - the hundred thousands digit - for it is the biggest number among numbers from 0-9. Furthermore, since there is no restriction as to whether the same number could or could not repeat, we can still use "9" all throughout the remaining 5 digits (ten thousands down to ones digit), giving the greatest possible 6-digit number, 999,999.
To check if the number we have obtained, 999,999, is divisible by 11, we need to use the divisibility rule for 11. Sum of even digits is 18 (9+9+9); sum of odd digits is 18 (9+9+9) as well. The difference of the sums is 0. Therefore, 999,999 is divisible by 11, hence the greatest 6-digit number multiple of 11.
As for the 6-digit number whose digits' sum is 40 and multiple of 11, we can assume its first 4 digits to be as 999,9** - four 9s is already equal to 36 - we only need to find two numbers whose sum is 4 to make the sum's total be 40. These remaining numbers should also be divisible by 11 since 9999 is already divisible by 11. The remaining numbers should only be 2 and 2 since the sum of its digit is 4, and that if we put them together, 22, a multiple of 11. This is the only combination of numbers that will give a sum of 4 while maintaining its structure a multiple of 11.
At last, the greatest 6-digit number whose digits' sum equal to 40 and is a multiple of 11 is 999,922. True, for 999,922 / 11 is equal to 90,902.
y= 2x + 1
Answers
Answers
Answer:
D. 7x + 2y < 2
Explanation:
The two general forma of the inequality are:
y < mx + c .........> In this case, shading is below the boundary line
y > mx + c .........> In this case, shading is above the boundary line
Since we are looking for the inequality with shading below the boundary line, therefore, we are looking for an y < mx + c format
Now, let's check the givens:
y - x > 5
Rearranging, we would get:
y > x + 5
The shading is above the boundary line. This option is incorrect
2x - 3y < 3
Rearranging, we would get:
2x - 3 < 3y
The shading is above the boundary line. This option is incorrect
2x - 3y < 7
Rearranging, we would get:
2x - 7 < 3y
The shading is above the boundary line. This option is incorrect
7x + 2y < 2
Rearranging, we would get:
2y < -7x + 2
The shading is below the boundary line. This option is correct
3x + 4y > 12
Rearranging, we would get:
4y > -3x + 12
The shading is above the boundary line. This option is incorrect
As a second solution, I attached the graphs of the 5 given functions.
Observing these graphs, we will find that the correct one is D
Hope this helps :)
D. 7x + 2y < 2
Explanation:
The two general forma of the inequality are:
y < mx + c .........> In this case, shading is below the boundary line
y > mx + c .........> In this case, shading is above the boundary line
Since we are looking for the inequality with shading below the boundary line, therefore, we are looking for an y < mx + c format
Now, let's check the givens:
y - x > 5
Rearranging, we would get:
y > x + 5
The shading is above the boundary line. This option is incorrect
2x - 3y < 3
Rearranging, we would get:
2x - 3 < 3y
The shading is above the boundary line. This option is incorrect
2x - 3y < 7
Rearranging, we would get:
2x - 7 < 3y
The shading is above the boundary line. This option is incorrect
7x + 2y < 2
Rearranging, we would get:
2y < -7x + 2
The shading is below the boundary line. This option is correct
3x + 4y > 12
Rearranging, we would get:
4y > -3x + 12
The shading is above the boundary line. This option is incorrect
As a second solution, I attached the graphs of the 5 given functions.
Observing these graphs, we will find that the correct one is D
Hope this helps :)
When you change the inequality as the formula of y>ax+b, it is over the boundary line. While the formula is y<ax+b, it is below the boundary line. So we can get the answer is D.
Draw 4 cards from a deck of cards, without replacement, and record each time the card is an ace.
Roll one die 10 times and record each time you roll and prime number.
Roll 2 number cubes until rolling a sum of 8 and record how many trials you completed.
Answers
Answer: the last option.
A binomial experiment is an experiment with only two possible outcomes.
In the first option, there is three possible outcomes (i.e. A, B, C)
In the second option, there is four poissible outcomes (i.e. ace, king, queen, jack)
The third option has six possible outcomes (i.e. six faces of a die)
In the last option, there are two possible outcomes (i.e. '2 number cubes')
A binomial experiment is an experiment with only two possible outcomes.
In the first option, there is three possible outcomes (i.e. A, B, C)
In the second option, there is four poissible outcomes (i.e. ace, king, queen, jack)
The third option has six possible outcomes (i.e. six faces of a die)
In the last option, there are two possible outcomes (i.e. '2 number cubes')
Answers
Remember Volume of a cube is V=s to the power 3. So in exponential form the answer to your question is V=(14 ft) to the power 3. Or without units. V=(14) to the power 3.
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