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kaheart [24]
2 years ago
7

Pls help me I really need help so I can not go through summer school pls

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
5 0
4 i think sorry if i’m wrong
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Can someone help me in this question?
igomit [66]

Answer:

3347000 cm is 33.47 km to cm

3347000/64=

401.92 is the circumference

since it is pi * d

do 3347000/401.92

8327.52786624 round to

8327 full revolutions

3 0
2 years ago
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Find the value of the trig function indicated
taurus [48]

Answer:

Step-by-step explanation:

base x height divided by 2

4x3=12

12 divided by 2 = 6

6x5= 30

6 0
3 years ago
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Explain why an equilateral polygon is not necessarily a regular polygon
poizon [28]
For a polygon to be regular it must have all sides the same length and all interior angles the same.
5 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
What is 93 times 10 to the sixth power
Rashid [163]
<span>93000000

...........................................</span>
8 0
3 years ago
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