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MA_775_DIABLO [31]
3 years ago
6

In the diagram, the radius of the outer circle is 2x cm and

Mathematics
1 answer:
hodyreva [135]3 years ago
8 0

Area shaded = Area big circle- Area of small circle;

200 pi= pi•(2x)^2 -pi•6^2;

200pi= pi•4x^2 -pi•36;

200pi=pi•4(x^2 -9) divide both sides by 4pi;

50=x^2 -9; So x=sqrt(59)~7.68cm

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What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

Doing this again and again, you'll see a similar pattern emerge:

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
3 years ago
What is the area of the shaded region? A. 21mm B. 24 mm C. 42 mm D. 48 mm
castortr0y [4]

Answer:

Can i see a pic of the shaded part?

Step-by-step explanation:

8 0
2 years ago
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8 0
3 years ago
34. A vessel is in the the form of a hemispherical bowl mounted by a hollow cylinder The diameter
Nostrana [21]

Answer:

V=20.85\ m^3

Step-by-step explanation:

Given that,

The vessel is in the the form of a hemispherical bowl mounted by a hollow cylinder.

The diameter  of the sphere is 3.5 m.

The height of the cylinder is 2 m

Height of the cylinder = Height of bowl− Radius of hemisphere

= 2 - (3.5/2)

= 0.25 m

The volume of the vessel is given by :

V = volume of hemispherical bowl + volume of cylinder

So,

V=\dfrac{2}{3}\pi r^3+\pi r^2 h\\\\V=\dfrac{2}{3}\times \dfrac{22}{7}\times (\dfrac{3.5}{2})^3+\dfrac{22}{7}\times (3.5)^2\times 0.25\\\\V=20.85\ m^3

So, the volume of the vessel is 20.85\ m^3.

5 0
2 years ago
4. Which set is an example of like fractions?
Andrew [12]
A, they both equal to 1 when simplified
6 0
3 years ago
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