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storchak [24]
3 years ago
9

Troy says that 0.9>0.90 because tenths are greater than hundredths.Keith says that 0.9<0.90 bra side 90 is greater than 9.

Is either Troy or Keith correct?Explain
Mathematics
2 answers:
stellarik [79]3 years ago
5 0

Answer:

Neither Troy nor Keith is correct.

Step-by-step explanation:

Neither Troy nor Keith is correct.

The value 0.9 is the same as the value 0.90.

The zeros after a decimal point are always insignificant unless they are followed by a number different from zero.

0.90 = 0.900 = 0.9000 = 0.90000 = ... = 0.9000000000

and all these can be easily written as 0.9, as the zeros after the decimal point do not change the value.

But 0.901, 0.9000004, 0.90002, and so on are all different, and the zeros are relevant, as they are followed by numbers different from zero.

earnstyle [38]3 years ago
5 0

Answer:

None of them is correct

0.9=0.90

Step-by-step explanation:

0.9>0.90 (not valid)

0.9<0.90 (not valid)

1. when dealing with decimals, the digits on the left hand side of the decimal point is a whole number.

The digits at the right hand side are considered individually.

2. the place value of the first digit is tenth, next is hundredth

3. if the last digits at the right hand side of the decimal point are zeros, then they are insignificant having no place value.

Hence, 0.9 is not greater 0.90

0.9 is not also less than 0.90,

because, the '0' at the right hand side of the decimal point, immediately after 9 is insignificant.

Therefore, 0.9=0.90

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

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