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3 years ago

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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.

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.

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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:

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2 years ago

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What dose y and x equal

natima [27]

X equals -2

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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,

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

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3 years ago

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B. 13

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2 years ago

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