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3 years ago

What is the 5th term of this geometric sequence? 100, 20, 4, 0.8, …

1 answer:

6 0

First we have to find out the common number of them.
For that purpose, the first number should be divided by next number.
100/20 = 5
20/4 = 5
4/0.8 = 5

we have found that 5 is the common number.
0.8/5 = 0.16

therefore the answer is c

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45six +13six <br><br> Carry out the addition shown using base six notation

Andrews [41]

Step-by-step explanation:

45 6

13 6

_____ +

02 6

________

6 /..8

..... -6

_______

..... 2

_______

6/5

...5

_____

8 0

2 years ago

Read 2 more answers

Identify √169/64 as rational or irrational and explain your reasoning.

Alex_Xolod [135]

Answer:

Rational

Step-by-step explanation:

This is rational because the square root of 169/64 would be 13/8 and any rational number could be put into a fraction

4 0

3 years ago

The ratio of the side lengths of a quadrilateral is 3:3:5:8, and the perimeter is 380cm. What is the measure of the longest side

castortr0y [4]

Answer:

160

Step-by-step explanation:

Add the ratios together to get the sum of them is 19. Since the perimeter is 380, divide 380 by 19 to get 20.

The shortest side is 3(20) = 60,

the next side is 5(20) = 100, and

the longest side is 8(20) = 160

3 0

3 years ago

Solve. Chanasia has 6.25 gallons of paint. She wants to use 2/5 of the paint to paint her living room. How many gallons of paint

deff fn [24]

Answer: 2.5 gallons

Step-by-step explanation:

Given that:

Gallons of paint had = 6.25 gallons

Amount needed to paint room = 2/5 of the paint

Gallons of paint needed :

2/5 of 6.25

(2 * 6.25) / 5 gallons

= 12.5 / 5

= 2.5 gallons

3 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago