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3 years ago

What is the square root of -1?

Mathematics

1 answer:

Westkost [7]3 years ago

3 0

Answer:

$\sqrt{i}$

Step-by-step explanation:

i = -1

This is a complex number

in here we take -1 = i

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lys-0071 [83]

Multiply the weight in tonnes by the conversion factor ( 1 tonne = 1000 kg) to get the weight in kg's:

170 tonnes x 1000 = 170,000

Now convert to scientific notation, by moving the decimal point to have only 1 digit to the left of the decimal.

To do this, you would need to move the decimal 5 places to the left to get 1.70000

Now remove the 0's and multiply the number by 10 raised to the number of paces the decimal was moved to get:

1.7 x 10^5 kg's.

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

What is the product of (x+2) (x-4)

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3 years ago

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3 years ago

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A conical water tank with vertex down has a radius of 13 feet at the top and is 21 feet high. If water flows into the tank at a

VLD [36.1K]

Answer:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

Step-by-step explanation:

We know that the conical water tank has a radius of 13 feet and is 21 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

$\frac{dV}{dt}=\frac{30\text{ ft}^3}{\text{min}}$

We want to find how fast the depth of the water is increasing when the water is 17 feet deep. So, we want to find dh/dt.

First, remember that the volume for a cone is given by the formula:

$V=\frac{1}{3}\pi r^2h$

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

$\frac{r}{h}=\frac{13}{21}$

Multiply both sides by h:

$r=\frac{13}{21}h$

So, let's substitute this in r:

$V=\frac{1}{3}\pi (\frac{13}{21}h)^2h$

Square:

$V=\frac{1}{3}\pi (\frac{169}{441}h^2)h$

Simplify:

$V=\frac{169}{1323}\pi h^3$

Now, let's take the derivative of both sides with respect to t:

$\frac{d}{dt}[V]=\frac{d}{dt}[\frac{169}{1323}\pi h^3}]$

Simplify:

$\frac{dV}{dt}=\frac{169}{1323}\pi \frac{d}{dt}[h^3}]$

Differentiate implicitly. This yields:

$\frac{dV}{dt}=\frac{169}{1323}\pi (3h^2)\frac{dh}{dt}$

We want to find dh/dt when the water is 17 feet deep. So, let's substitute 17 for h. Also, let's substitute 30 for dV/dt. This yields:

$30=\frac{169}{1323}\pi (3(17)^2)\frac{dh}{dt}$

Evaluate:

$30=\frac{146523}{1323}\pi( \frac{dh}{dt})$

Multiply both sides by 1323:

$39690=146523\pi\frac{dh}{dt}$

Solve for dh/dt:

$\frac{dh}{dt}=\frac{39690}{146523}\pi$

Use a calculator. So:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

The water is rising at a rate of approximately 0.086 feet per minute.

And we're done!

Edit: Forgot the picture :)

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3 years ago