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Daniel [21]
3 years ago
6

Please help It’s due at 11:59 it’s 9:29 right now!!

Mathematics
1 answer:
yan [13]3 years ago
3 0

Step-by-step explanation:

  1. 8969kg/cm^3
  2. 20oz jar of$2.78
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Greg spent 6.4 hours working with his dad one Saturday. If his dad paid him$45.76, how much did Greg make per hour?
nordsb [41]

Answer:

$7.15

Step-by-step explanation:

The best way to solve this is by crossing out the data you don't need so you're left looking at the key pieces to put the equation together. If his dad paid him $45.76 for 6.4 hours, then all you need to do is divide 45.76 by 6.4. (Hint: Use a calculator, it makes it ten times faster than trying to write it out!)

45.76 ÷ 6.4 = 7.15

So Greg made $7.15 an hour working with his dad.

6 0
3 years ago
Which expression can be used to check the answer to 56 ÷ (-14) = n?
kozerog [31]
A. Multiplication is the ONLY function that reverses what division did. If you did 56/-14 = n, then n x -14 should equal 56.
6 0
3 years ago
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1. Lựa chọn yêu cầu cần đạt trong Chương trình Giáo dục phổ thông môn Toán 2018 lớp 3, xây dựng một công cụ sử dụng trong đánh g
maks197457 [2]

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So basically you add x with 9 and you get 9x

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3 years ago
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4 0
2 years ago
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The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
jeyben [28]

Answer:

The period of Y increases by a factor of k^ {3/2} with respect to the period of X

Step-by-step explanation:

The equation T ^ 2 = a ^ 3 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:  

T_y ^ 2 = a_y ^ 3

For planet X:  

T_x ^ 2 = a_x ^ 3

To know the factor of aumeto we compared T_x with T_y

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:  

a_y ^ 3 = (a_xk) ^ 3

So

\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x

Then, the period of Y increases by a factor of k^ {3/2} with respect to the period of X



4 0
3 years ago
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