The equation 7^2=a^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is k times the mean distance from the sun as planet X, by what factor is the orbital period increased?
Answers
Answer:
The period of Y increases by a factor of with respect to the period of X
Step-by-step explanation:
The equation shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?
For the planet Y:
For planet X:
To know the factor of aumeto we compared with
We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:
Finally the period of Y increases by a factor of with respect to the period of X
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True or false:Every quadrilateral is a square. Every square is a quadrilateral. Every rhombus with four right angles is a square.
Answers
Answer:
2 grams
Step-by-step explanation:
multiply by 1000
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The number 15 times 3 is equals 45.
What is Multiplication?
To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.
Now, We know that;
⇒ 15 × 3
⇒ 45
Hence, 15 times 3 is equals 45.
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hope I help.
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3 to the 2nd is 9
3 to the 3rd is 27
3 to the 4th is 81
3 to the 5th is 243
3 to the 6th is 729
Answers
Answer:10
Step-by-step explanation:
Final answer:
The cost of each adult ticket is $28, but there is no solution for the cost of each student ticket in this case.
Explanation:
To solve this problem, we can set up a system of equations using the given information. Let's represent the cost of an adult ticket as 'a' and the cost of a student ticket as 's'. From the first equation, we know that 4a + 2s = 64. From the second equation, we know that 3a + 3s = 60. Now we can solve this system of equations.
Multiplying both sides of the second equation by 2 gives us 6a + 6s = 120. We can subtract the first equation from this equation to eliminate 's'.
6a + 6s - (4a + 2s) = 120 - 64
Expanding and simplifying the equation gives us 2a = 56. Dividing both sides by 2, we find that a = 28. Now we can substitute this value into either of the original equations to find 's'.
Using the first equation: 4(28) + 2s = 64
Simplifying the equation gives us 112 + 2s = 64
Subtracting 112 from both sides gives us 2s = -48. Dividing both sides by 2, we find that s = -24.
However, since we're talking about the cost of tickets, we can't have a negative value. Therefore, there is no solution for 's' in this case.
The cost of each adult ticket is $28, but there is no solution for the cost of each student ticket in this case.
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arranged in 3 equal columns. How many
rows of solar panels are on this house?
Explain how to solve the problem.
Answers
Answer:
There are 6 rows of solar panels on this house.
Step-by-step explanation:
18/3=6
Answer:
6
Step-by-step explanation:
18 divided equally into 3 columns should give you 6 rows.
18 ÷ 3 = 6