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Tanya [424]
3 years ago
10

G=7/5(p-45) The solution is p=__g+___ Simplify your answers. Type an integer or fraction.

Mathematics
2 answers:
amm18123 years ago
7 0

So firstly, multiply both sides by 5/7 (reciprocal of 7/5): \frac{5}{7}g=p-45

Next, add both sides by 45 and your answer will be: \frac{5}{7}g+45=p

iVinArrow [24]3 years ago
7 0

\frac{7}{5}(p-45)=g \implies \\ p-45 =\frac{5}{7}g= \implies\\ p=\frac{5}{7}g+45

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The coordinates of the points below represents the vertics of a rectangle
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Not sure what the question is, but I'm guessing you want to find out the area of the rectangle.

The distance between points P and Q is 4.

The distance between points Q and R is 3.

So the area is 3 * 4 = 12.

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3 years ago
Maple town is 16 kilometers due north of the airport and sachet town is due east of the airport. The shortest distance between m
Alexxandr [17]

Answer:

Sachet town is 12km from the airport

Step-by-step explanation:

In this question, we are asked to calculate the distance from sachet town to the airport.

This can be calculated by referring to a diagrammatic representation. Please check attached file.

Let’s proceed with the calculations however.

From the question, we can see that we have formed a Right angled triangle from the diagram and we need to calculate the distance x.

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x^2 + 16^2 = 20^2

x^2 = 20^2 - 16^2

x^2 = 400 - 256

x^2 = 144

x = sq.rt of 144

x = 12km

6 0
3 years ago
Read 2 more answers
The answer please answer
strojnjashka [21]
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5 0
3 years ago
What is the value of 5/6 divided by 3/7
vaieri [72.5K]
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4 0
3 years ago
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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