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2 years ago

Why is this not working -2^2

Mathematics

2 answers:

BARSIC [14]2 years ago

5 0

Answer:

-4

Step-by-step explanation:

(-2)^2=4 because a negative number squared is always positive

-(2)^2=-4

Alex17521 [72]2 years ago

4 0

Answer:

try putting the minus value in brackets

Step-by-step explanation:

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GIVING 30 POINTS

FromTheMoon [43]

Answer:

A. 502.4 cm^3

Step-by-step explanation:

r=8/2=4 cm

V=pi*r^2*h=3.14*16*10=502.4 cm^3 (A)

7 0

3 years ago

Read 2 more answers

Cloverdale Middle School is buying new helmets for the football team. Each helmet costs $89.95. How much money does the school s

seraphim [82]

Answer:

4047.75$

Step-by-step explanation:

4 0

2 years ago

Amelia has 3 gallons of paint .Mason have 10 quarts .How many more pints of paint does AMELIA have then mason.

Ber [7]

Answer:

4 Pints

Step-by-step explanation:

According to the scenario, calculations are as follows,

As we know, 1 gallon = 8 Pints

1 Quarts = 2 Pints

So, Amelia = 3 Gallons = 3 × 8 Pints = 24 Pints

Mason = 10 quarts = 10 × 2 Pints = 20 Pints

So, Number of more pints Amelia has = Amelia number of pints - Mason number of pints

= 24 pints - 20 pints

= 4 Pints

6 0

3 years ago

I have ten minutes to send it in.

dsp73

Refer to the attachment for answer

3 0

2 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago