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3 years ago

5

What is the decimal expansion of the following fraction? 1/8 A. 0.125 B. 0.125 C. 1.8 D. 0.18

1 answer:

qaws [65]3 years ago

3 0

Answer:

0.125....................

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A rate of 4 centimeters per second is equal to how many meters per hour?

vampirchik [111]

So, let's start with the basics.
One meter is 100 centimeters. Seeing that we have 60 seconds in a minute, and 60 minutes in an hour. The formula would look like this.
(60*60)4
3600*4 = 14400 centimeters
So then we divide it by 100, and
144 meters

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3 years ago

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3 years ago

Read 2 more answers

Select the correct answer.

meriva

Answer:

D, x = (ln(0.5) - 1)/2

Step-by-step explanation:

8e^(2x+1) = 4

e^(2x+1) = 0.5

2x+1 = ln(0.5)

2x = ln(0.5) - 1

x = (ln(0.5) - 1)/2

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3 years ago

PLEASE ILL MAKE BRAINIEST!!!

expeople1 [14]

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$ , $AB = 12$ , $BC = 15$ , $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$ , $EF = EC + CF$ , $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$ :

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$ , $BC = 15$ and $CD = 12$ , then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.

4 0

3 years ago

What is the equation of the trend line?

dimulka [17.4K]

N=(x)+M+(B) that's ur answer

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4 years ago