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3 years ago

What is 59÷81,748 in math

Mathematics

1 answer:

Zinaida [17]3 years ago

4 0

I did The maTh and iTs 1,385.5.

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Given that (-4,2) is on the graph of f(x) find the corresponding point for the function -1/2f(x)

Aleks [24]

Answer:

The corresponding point is (-4.-1)

Step-by-step explanation:

we know that

The point (-4,2) is on the graph of f(x)

For x=-4

f(x)=2

therefore

For x=-4

-(1/2)f(x)=-(1/2)*2=-1

The corresponding point is (-4.-1)

8 0

2 years ago

X f(x)

Montano1993 [528]

I would say A hope I helped

8 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago

The burning time of a very large candle is normally distributed with mean of 2500 hours and standard deviation of 20 hours. Find

dezoksy [38]

Answer: C.-1.5

Step-by-step explanation:

Given: The burning time of a very large candle is normally distributed with mean $(\mu)$ of 2500 hours and standard deviation $(\sigma)$ of 20 hours.

Let X be a random variable that represent the burning time of a very large candle.

Formula: $Z=\dfrac{X-\mu}{\sigma}$

For X = 2470

$Z=\dfrac{2470-2500}{20}\\\\\Rightarrow\ Z=\dfrac{-30}{20}\\\\\Rightarrow\ Z=\dfrac{-3}{2}\\\\\Rightarrow\ Z=-1.5$

So, the z-score they corresponds to a lifespan of 2470 hours. =-1.5

Hence, the correct option is C.-1.5.

7 0

2 years ago

A rugby match lasts for 80 mins. 20 mins of a match have been played. a) what Fraction of he match has been played? b) what perc

Delicious77 [7]

25% has been played and 1/4

8 0

1 year ago

Read 2 more answers