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2 years ago

What is the common ratio between successive terms in the sequence?

Mathematics

2 answers:

GrogVix [38]2 years ago

4 0

Answer:

It's 1/3.

Step-by-step explanation:

Common ratio = 9/27 = 1/3.

3 / 9 = 1/3

1 / 3 = 1/3 and so on.

Assoli18 [71]2 years ago

4 0

Answer:

r = $\frac{1}{3}$

Step-by-step explanation:

The ratio between successive terms is

$\frac{9}{27}$ = $\frac{1}{3}$

$\frac{3}{9}$ = $\frac{1}{3}$

$\frac{1}{3}$

$\frac{\frac{1}{3} }{1}$ = $\frac{1}{3}$

$\frac{\frac{1}{9} }{\frac{1}{3} }$ = $\frac{1}{3}$

$\frac{\frac{1}{27} }{\frac{1}{9} }$ = $\frac{1}{3}$

Since the common ratio between terms is constant then sequence is geometric

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a man divides ₹90,000 among wife son and daughter in the ratio 1/3:1/4:1/6 . find amount received them

Pani-rosa [81]

The lowest common multiple of 3,4 and 6 is 12. Therefore if you multiply the ratio by 12 you get 12/3 : 12/4 : 12/6 which is equal to 4:3:2.
When you add 4,3 and 2 the answer is 9 therefore the £90,000 is split into 9 parts. £90,000 divided by 9 = £10,000 which means each part is £10,000.
The wife receives 4 parts so she receives 4 x £10,000 = £40,000
The son receives 3 parts so he receives £30,000
The daughter receives 2 parts so she receives £20,000

4 0

2 years ago

Add 17/99 + 27/99 + 15/99

Helen [10]

Answer:

59/99

Step-by-step explanation:

Since the denominator is the same you add the numerator values for each fraction 27+17+15 = 59

8 0

2 years ago

Read 2 more answers

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0

3 years ago

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Maru [420]

Answer:46%

Step-by-step explanation:

5 0

3 years ago

Whats the answer?<br> And step by step?

s344n2d4d5 [400]

Answer:

one answer

Step-by-step explanation:

set the equal to each other and you get one answer

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7 0

3 years ago