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Lady bird [3.3K]

3 years ago

5

Estimate 200 times 3

2 answers:

Sonja [21]3 years ago

7 0

I "estimate" 600....?

likoan [24]3 years ago

6 0

Answer

200*3= 600

Step-by-step explanation:

hope this helps.

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Mai travels 90 miles in 2 hours by boat. At this rate,how far does Mai go in 3 hours?

Marysya12 [62]

Answer:

135 miles in 3 hours

Step-by-step explanation:

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3 years ago

John Doe estimates that the price elasticity of demand for new SUVs to be 0.75. If the price of SUVs rose by 15%; would the quan

Charra [1.4K]

If the coefficient of demand for the SUV is 0.75 this means that it has a relatively inelastic demand since it is less than. In other words, there is only a little alteration in demand when prices change.
So when the price of SUV rise by 15%, and it has a coefficient of 0.75, we can anticipate only 11.25% decrease in its demand. This is for the reason that the SUVs do not have many substitutes for it.
So to solve:
(x/15%) = 0.75
Then simply solve for x:
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4 0

3 years ago

Write an equation reprezeting x axis

Rudiy27

Answer:

y=0

Step-by-step explanation:

All y-values along the x-axis are equal to 0.

3 0

3 years ago

Need help with simplifying expressions

Ilia_Sergeevich [38]

Answer:

A) -20

Step-by-step explanation:

distribute the 4y and the -4 to the parentheses

16y^2 -16y+16y-20-16y^2

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Mark brainliest??

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3 years ago

Read 2 more answers

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

4 years ago