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3 years ago

What is 1/2,0.55,5/7 in least to greatest?

Mathematics

2 answers:

olga_2 [115]3 years ago

6 0

Answer:

5/7, 1/2, 0.55

I believe this is the correct answer. Correct me if i'm wrong.

Neporo4naja [7]3 years ago

3 0

Answer:

1/2, 0.55, 5/7 is your answer

Step-by-step explanation:

Change each into decimal

1/2 = 0.50

0.55 = 0.55

5/7 = ~0.71

0.50 < 0.55 < 0.71

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The range of the function f(x) = x + 5 is {7, 9). What is the function's domain?

Maru [420]

Answer:

(2,4)

Step-by-step explanation:

The range is the y values

y = x+5

Let y=7

7 =x+5

Subtract 5

7-5 x+5-5

2=x

Let y= 9

Subtract 5

9-5 x+5-5

4=x

The domain is (2,4)

8 0

3 years ago

The dimensions in inches of a shipping box at we ship 4 you can be expressed as width x, length x+5, and height 3x-1. The volume

MakcuM [25]

Answer:

Connexus: C,D,C,A,A.

Step-by-step explanation:

6 0

3 years ago

A number cube is rolled 350 times and these are the results: 60 ones, 65 twos, 54 threes, 63 fours, 62 fives, and 46 sixes. What

BlackZzzverrR [31]

The experimental probability can be represented as a fraction of the number of times a five was not the result over the total number of times the cube was rolled.

The answer would be 288/350 or 144/175.

7 0

3 years ago

A train on a straight track goes in the positive direction for 6.7 km , and then backs up for 4.0 km . Is the distance covered b

RSB [31]

Answer:

Distance is greater than the displacement.

Step-by-step explanation:

We are given that a train on a straight track goes in the positive direction.

We have to find that distance covered by the train is greater than, lees than or equal to its displacement.

Train covered distance in positive direction=6.7 km

Train covered distance in back=4.0 km

Distance covered by the train=6.7+4=10.7 Km

Because distance is the length of path covered by object.

Displacement =Final position -Initial position

Displacement =6.7-4=2.7 km

10.7 > 2.7

Distance > Displacement

Hence, Distance is greater than the displacement.

7 0

3 years ago

Please help me answer this question

ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

$tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}$

Then,

$\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}$

$=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}$

$=\frac{2e^{-x}}{2e^{x}}$

$=\frac{e^{-x}}{e^{x}}$
$=e^{-x} \times \frac{1}{e^{x}}$

$= e^{-x} \times e^{-x}$

$= e^{-x+-x}$

$= e^{-x-x}$

$= e^{-2x}$

Hence, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Learn more on Proof of Identities here: brainly.com/question/2561079

#SPJ1

6 0

2 years ago