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3 years ago

5

The ratio of your monthly allowance to your friend’s monthly allowance is 5 : 3. If your friend received a monthly allowance of

$51, how much did you receive?

1 answer:

Sever21 [200]3 years ago

5 0

Answer:

i recience 10 dollars

Step-by-step explanation:

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The middle school is 2.72 km from marshes House and 1.52 km from Ryan’s house how much further does Martha live from the middle

Alex_Xolod [135]

1.20kilometers farther away from Ryan’s house

6 0

3 years ago

⅜ of the animals at the zoo equals to 600. 40% of all the animals at the

raketka [301]

Answer:

960

Step-by-step explanation:

3/8*total number of animals = 600

total number of animals = 8/3 * 600 = 1600

number of animals fed = 40/100 * 1600 = 640

animals still to be fed = 1600 - 640 = 960

5 0

3 years ago

A certain forest covers an area of 4800 km^2 . Suppose that each year this area decreases by 5.25% . What will the area be after

9966 [12]

Answer:

the area after 6 years is 3,473 km^2

Step-by-step explanation:

The computation of the area after 6 years is as follows:

= Area × (1 - decreased percentage)^number of years

= 4,800 km^2 × (1 - 5.25%)^6

= 4,800 km^2 × 0.9475^6

= 3,473 km^2

Hence, the area after 6 years is 3,473 km^2

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

I need help with this question

Amiraneli [1.4K]

Answer:

what is the question be more specific

7 0

3 years ago

Read 2 more answers