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3 years ago

Thirty-eight and ninety-seven hundredths kilograms in standard form

Mathematics

2 answers:

Galina-37 [17]3 years ago

4 0

Answer:

The answer is 38.97 kilograms.

Step-by-step explanation:

Thirty-eight and ninety-seven hundredths kilograms in standard form is 38.97 kilograms.

Thirty eight is simply 38.

And ninety-seven hundredths means that 97 is written after decimal. This is because when there exists an 'and' in a mathematical number, this means there is a decimal.

So, the answer is 38.97 kilograms.

Fynjy0 [20]3 years ago

3 0

Standard form - Regular typing and writing with numbers

so it's 38.97 kg

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Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is

OLga [1]

Answer:

reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda.

Step-by-step explanation:

I hope this helps, I'm in 11th grade and I'm pretty sure the answer is correct. Good Luck!!!

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3 years ago

Expanding logarithmic Expression In Exercise,Use the properties of logarithms to rewrite the expression as a sum,difference,or m

wel

Answer:

$\frac{1}{2}ln(2x)-\frac{1}{2}ln(x^2-1)$

Step-by-step explanation:

In 2x/(x^2 - 1)1/2

$ln(\frac{2x}{x^2-1} )^\frac{1}{2}$

Apply the property of natural log

ln x^m = m ln(x) move the exponent before ln

$ln(\frac{2x}{x^2-1} )^\frac{1}{2}$

$\frac{1}{2}ln(\frac{2x}{x^2-1})$

ln(m/n)= ln m - ln n

$\frac{1}{2}(ln(2x)-ln(x^2-1))$

multiply 1/2 inside the terms

$\frac{1}{2}ln(2x)-\frac{1}{2}ln(x^2-1)$

6 0

2 years ago

consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

a)Generally the concave upward interval X is mathematically given as

$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

5 0

2 years ago

Can someone help me please.

bazaltina [42]

The answer is B. And I am not that sure about my answer, but it has to be the correct answer for what you put. THANK YOU!!!!

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3 years ago

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astraxan [27]

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3 years ago