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2 years ago

Need answer important pls

Mathematics

1 answer:

Natali5045456 [20]2 years ago

7 0

Answer:

C {-2 1/6, 1 5/6}

Step-by-step explanation:

This will solve your answer

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The infinite geometric sum formula can be used to write 0.126126126...as a fraction. What is the numerator of this reduced fract

Kryger [21]

Answer:

B. 14

Step-by-step explanation:

The sum of infinite geometric series is given by

$\frac{a}{1-r}$

0.126126126... = $\frac{126}{1000}+\frac{126}{1000000}+\frac{126}{1000000000}$

Here a =126/1000, r= 1/1000

0.126126126... = $\frac{126/1000}{1-1/1000}$ =126/999 = 14/111

∴ 0.126126126... = 14/111

So, the numerator of the reduced fraction is 14.

8 0

4 years ago

Read 2 more answers

If George is 33 1/3% richer than Pete, than Pete is what percent poorer than George?

OleMash [197]

Answer:

25%

Step-by-step explanation:

George is 33 $\frac{1}{3}$ % ( $\frac{100}{3}$ %) richer than Pete. Let Pete's percentage of wealth be 100%.

Thus George percentage of wealth = 100% + $\frac{100}{3}$ %

= $\frac{400}{3}$ %

= 133 $\frac{1}{3}$ %

Pete's percent poorer than George can be determined by;

= $(\frac{100}{3})$ ÷ $(\frac{400}{3} )$ × 100

= $(\frac{100}{3})$ × $\frac{3}{400}$ ×100

= 0.25 × 100

= 25%

Pete is 25% poorer than George.

3 0

3 years ago

Inverse laplace of L^-1 {5s/s^2 + 3s - 4}

Salsk061 [2.6K]

Answer:

$e^t+e^{-4t}$

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

$\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\$

$From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}$