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Shkiper50 [21]
3 years ago
14

The latitude value of Toronto is 43.70 degrees, and the longitude value is 79.40 degrees. The latitude value of Melbourne is -37

.81 degrees, and the longitude value is 144.96 degrees. The two cities are degrees apart in latitude. The two cities are degrees apart in longitude.
Mathematics
1 answer:
Aneli [31]3 years ago
3 0
I'm hazarding a guess it's 224.36
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lamont cotton is 4% comission on the first 10,000 of monthly sales mens 10% on all sales over 10,000. last month his 38,00. what
Pani-rosa [81]

Given:

Lamont cotton is 4% commission on the first 10,000 of monthly sales and 10% on all sales over 10,000.

Consider his last month sales is 38,000.

To find:

The total commission.

Solution:

4% commission on the first 10,000 of monthly sales is

10,000\dfrac{4}{100}=400

Sales over 10,000 is

38,000-10,000=28,000

10% commission on all sales over 10,000 is

28,000\times \dfrac{10}{100}=2,800

Now, the total commission is

400+2,800=3,200

Therefore, the total commission is 3,200.

8 0
2 years ago
1.25 is closer to 1.04 or not ?<br> plz heelp
Ray Of Light [21]

Answer:

Yes 1.25 is closer to 1.04

Step-by-step explanation:

Hope this helps!

3 0
2 years ago
Read 2 more answers
What dose y and x equal
natima [27]
X equals -2

Y equals 5
4 0
3 years ago
Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

6 0
3 years ago
Please answer thxxxx
Llana [10]
B. 13

12^2 + 5^2 = c^2
169 = c^2
13 = c


please mark as brainliest :D
4 0
2 years ago
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