MATHEMATICS HIGH SCHOOL

A merry-go-around horse is traveling at 10 feet per second when the merry go around is making 6 revolutions per minute. How far is the horse from the center of the merry-go-around?Please help

Answers

Answer 1

Answer:

Answer:

3,600

Step-by-step explanation:

$10 * 60 * 6$

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The estimated snowfall for a local city is 6.75 inches. The actual snowfall was 10.25 inches. What is the percent error for the amount of snowfall

Answers

Answer: ~ 34.15%

Step-by-step explanation:

To calculate the percent error for the amount of snowfall, you can use the following formula:

Percent Error = [(|Estimated Value - Actual Value|) / Actual Value] * 100%

In this case, the estimated snowfall is 6.75 inches, and the actual snowfall is 10.25 inches. Plugging these values into the formula:

Percent Error = [(|6.75 - 10.25|) / 10.25] * 100%

Percent Error = [(3.5) / 10.25] * 100%

Percent Error ≈ 34.15%

So, the percent error for the amount of snowfall is approximately 34.15%. This means that the estimated snowfall was about 34.15% lower than the actual snowfall.

Which number is lower?
A) 4.8
b) -9.9

Answers

-9.9 it’s that because it’s negative

Answer:

b) -9.9

Step-by-step explanation:

since its a negative its automatically smaller than any whole

hope this helps plz give brainliest

P(A) = 0.44 What are the odds for A?

Answers

$P(A)=0.44=(44)/(100)=(11)/(25)\n\n11\ odds\ on\ 25$

0,44 = 44 / 100

44 odds on 100

How many grams are in 15 kilograms? 1.5 150 1,500 15,000

Answers

Answer:

The last option 15,000 grams is correct.

Step-by-step explanation:

How many grams are in 15 kilograms?

We must know the relation between kilograms and grams.

1 kg = 1000 grams.

Now we have been asked to calculate the grams in 15 kilograms.

So, 15 kilograms has = $15*1000=15000$ grams

Therefore, the last option 15,000 grams is correct.

15000 because 15kg x 1000 = 15000g

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))aka: $(1-x^(sin(x)))/(x*log(x))$

Please include steps/explanation.

Answers

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_( x \to 0 ) ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_(x \to 0^(+)) (1- x^(\sin x) )/(x \log x ) \n~\n \large = \lim_(x \to 0^(+)) (1- x^(\sin x) )/( \log( x^x) ) \n~\n \large = \lim_(x \to 0^(+)) (1- x^(x) )/( \log( x^x) ) ~~ \normalsize{\text{ substituting x for sin x } } \n~\n \large = (\lim_(x \to 0^(+)) (1) - \lim_(x \to 0^(+)) \left( x^(x)\right) )/( \log( \lim_(x \to 0^(+))x^x) ) = (1-1)/(\log(1)) = (0)/(0)$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

$\large \lim_(x \to 0^(+)) (1- x^(x) )/( \log( x^x) ) =_(LH) \lim_(x \to 0^(+)) (0 -x^x( 1 + \log (x)) )/(1 + \log (x) ) \n = \large \lim_(x \to 0^(+)) (-x^x) = \large - \lim_(x \to 0^(+)) (x^x) = -1$

Therefore,

$\large \lim_(x \to 0^(+)) (1- x^(\sin x) )/(x \log x ) =\boxed{ -1}$

1/5 times 3/4 times 5/7 in simplest form