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kipiarov [429]
3 years ago
8

Find the unit rate. Round to the nearest hundredth, if necessary.

Mathematics
2 answers:
nekit [7.7K]3 years ago
5 0

Answer:

Usually, I would divide 300 and 17 to get my answer

Step-by-step explanation:

babymother [125]3 years ago
4 0

Answer:

17.65

Step-by-step explanation:

300/17

17.64705882352941

Round to hundred

17.65

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Dy/dx of ln(x/x^²+1)
exis [7]

\dfrac{dy}{dx}\ \ln\left(\dfrac{x}{x^2+1}\right)=\dfrac{1}{\frac{x}{x^2+1}}\cdot\dfrac{x'(x^2+1)-x(x^2+1)'}{(x^2+1)^2}\\\\=\dfrac{x^2+1}{x}\cdot\dfrac{1(x^2+1)-x(2x)}{(x^2+1)^2}=\dfrac{1}{x}\cdot\dfrac{x^2+1-2x^2}{x^2+1}=\dfrac{1-x^2}{x^3+x}\\\\Used:\\\\(\ln(x))'=\dfrac{1}{x}\\\\\left[f(g(x))\right]'=f'(g(x))\cdot g'(x)\\\\\left(\dfrac{f(x)}{g(x)}\right)'=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}

5 0
3 years ago
How are determinants used to find areas of polygons?
ANTONII [103]
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For example, consider the vertices of a triangle: (1,1), (2,3), (3,-1). Its area can be computed as
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4 0
3 years ago
A 12​-tooth gear on a motor shaft drives a larger gear having 42 teeth. If the motor shaft rotates at 700 ​rpm, what is the spee
nata0808 [166]

Answer:

203 rpm

Step-by-step explanation:

The speed of the larger gear can be calculated using the following equation:

v = \omega*R  

<u>Where</u>:

ω: is the angular velocity of the motor = 700 rpm

R: is the gear ratio

The gear ratio is the following:

R = \frac{n_{(a)}}{n_{(b)}}  

<u>Where:</u>

n(a): is the number of teeth on the small gear = 12 teeth  

n(b): is the number of teeth on the larger gear = 42 teeth

The gear ratio is:

R = \frac{12}{42} = 0.29

Now, the speed of the larger gear is:

v = \omega*R = 700 rpm*0.29 = 203 rpm

Therefore, the speed of the larger gear is 203 rpm.

I hope it helps you!                        

3 0
3 years ago
Express 4,400 in scientific notation
trapecia [35]
4.4 * 10^{3} 
4 0
4 years ago
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Jacqueline's piggy bank contains $6.20 in dimes and quarters. There are 32 coins in all. How many of each kind are there?
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20 quarters and 2 dimes!! (Take a look at my question if possible!?)
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