Answer:
x = 0
Step-by-step explanation:
Answer:
8x^2
Explanation:
First, do prime factorization of each of the coefficients:
32 ⇒ 2^5
24 ⇒ 2^3, 3
The greatest common factor (GCF) of the coefficients is 2^3 = 8.
Next, find the GCF of the variables:
x^2
x^2, y
The GCF of the variables is x^2.
Finally, multiply the GCF of the coefficients by the GCF of the variables to get:
8x^2
So we can setup variable equations for this:
so we're given that the length is 22feet longer than the width!
so as an expression length is equal to 22+w
and width as an expression is just w
so the formula for the perimeter for a rectangle is 2l+2w=perimeter
so fill it in now!
1616=2(22+w)+2w
1616=44+2w+2w
1616=44+4w
1572=4w
divide by four..
w=393
and length is simply 22 more than that as stated in the question!
so length is 415
dimensions are: 415x393
False...it does not include 3....it includes only numbers less then 3.
For it to include 3, u would need an equal sign in there..
Y < = 3...(thats less then or equal)...this would include 3
Answer:
- arc second of longitude: 75.322 ft
- arc second of latitude: 101.355 ft
Explanation:
The circumference of the earth at the given radius is ...
2π(20,906,000 ft) ≈ 131,356,272 ft
If that circumference represents 360°, as it does for latitude, then we can find the length of an arc-second by dividing by the number of arc-seconds in 360°. That number is ...
(360°/circle)×(60 min/°)×(60 sec/min) = 1,296,000 sec/circle
Then one arc-second is
(131,356,272 ft/circle)/(1,296,000 sec/circle) = 101.355 ft/arc-second
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Each degree of latitude has the same spacing as every other degree of latitude everywhere. So, this distance is the length of one arc-second of latitude: 101.355 ft.
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<em>Comment on these distance measures</em>
We consider the Earth to have a spherical shape for this problem. It is worth noting that the measure of one degree of latitude is almost exactly 1 nautical mile--an easy relationship to remember.
