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madam [21]
1 year ago
6

1. Convert the numeral to a numeral in base ten. 3205 six 43 125 4325 725

Mathematics
1 answer:
mario62 [17]1 year ago
8 0

Expanding the digits, we have

3205_6 \equiv 3\cdot6^3 + 2\cdot6^2 + 0\cdot6^1 + 5\cdot6^0 = \boxed{725}

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Jeff writes comic strips for local newspaper. The number of comic strips he created is represented by the function f(x)=3x, wher
grin007 [14]
If Jeff writes comic strips at a rate of 3 comics an hour then you must put 3,4,7&9 into the function and add up the products.
9+12+21+27=69
5 0
3 years ago
The difference between two natural no. is 196 and the ratio of the two no. is 9:5. Find Two
stellarik [79]

Answer:

441 and 295

Step-by-step explanation:

Let the natural number be 9x and 5x.

9x-5x=4x=196. x=49

7 0
3 years ago
Solve the system.
patriot [66]

Answer:

D.(2,-1,-1)

Step-by-step explanation:

-a+4b+2c=-8

3a+b-4c=9

b = -1

Substitute b=-1 into the other equations

-a -4 +2c = -8

3a -1 -4c =9

Multiply the first equation by 2 so we can eliminate c

-2a -8 +4c = -16

Add this to the second equation

-2a -8 +4c = -16

3a -1 -4c =9

-----------------

a -9 = -7

Add 9 to each side

a-9+9 = -7+9

a =2

b=-1

Now we need to find c

-a +4b +2c = -8

-2 -4 +2c = -8

-6+2c = -8

Add 6 to each side

-6+6 +2c = -8+6

2c = -2

Divide by 2

2c/2 = -2/2

c=-1

(2,-1,-1)

5 0
4 years ago
(2x^2-33x+16) divided by (x-16)
jolli1 [7]
I took me a while I got it
it's 2x−1
5 0
4 years ago
Consider the sequence {an} = {nrn}. Decide whether {an} converges for each value of r.
Rudiy27

Answer:

a). converges

b). diverges

c). converges

Step-by-step explanation:

{ $ a_n $ } = { $ nr^n $ }

Using radio test,

L = $ \lim_{n \rightarrow \infty} |\frac{a_n + 1}{a_n}| $

   = $ \lim_{n \rightarrow \infty} |\frac{(n+1)r^{n+1}}{nr^n}| $

   = $ \lim_{n \rightarrow \infty} |(1+\frac{1}{n})^r| $

   = $ \lim_{n \rightarrow \infty} |r| $

   = |r|

Therefore, $a_n$ converges in |r| < 1

a). r = 1/5

    $\{ a_n \}=  \{\frac{1}{5}, n \}$

This sequence is monotonically decreasing and bounded.

0 < $ a_n $ < 1

Hence, { $ a_n $ } converges.

b). r = 1

    { $ a_n $ } = { n }

This sequence is monotonically increasing sequence which is not bounded.

Hence, { $ a_n $ } diverges.

c). r = 1/6

$\{ a_n \}=  \{\frac{1}{6}, n \}$

This sequence is monotonically decreasing and bounded.

0 < $ a_n $ < 1

Hence, { $ a_n $ } converges.

For  |r| < 1, the $a_n$ converges.

3 0
3 years ago
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