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3 years ago

Where stocks are traded can be called any of these except ____. A. investment arena B. trading floor C. exchange D. market

Mathematics

2 answers:

vova2212 [387]3 years ago

8 0

Answer:

Option A is correct - investment arena

Step-by-step explanation:

Where stocks are traded can be called any of these except - investment arena.

We can find stocks trading in the trading floor, stock exchange and stock market but not in investment arena. Investment arena is no physical place.

Marina CMI [18]3 years ago

6 0

Answer: A. investment arena

Step-by-step explanation:

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3 years ago

Which of the following graphs represents the equation below?

Musya8 [376]

Answer:

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Step-by-step explanation:

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3 years ago

Dylan and Spencer went looking for Dylan's lost dog. Dylan traveled 100 meters to the neighbor’s house, 1.5 kilometers to the pe

pickupchik [31]

So Dylan traveled a total of
100 meters (to the neighbors house)
plus 1.5 kilometers which a kilometer means thousand meters so its really 1500 meters
plus 310 meters

For Dylans distance we can do 100 + 1500 + 310 = 1910
Specers distance is given which is 1269

To find the final answer we can do Dylans distance minus Spencers so 1910 - 1269 which equals 641.

The correct answer would be that Dylan traveled 641 meters farther than Spencer

Answer: 641

Hope this helps!

4 0

3 years ago

Read 2 more answers

Which equation has the solutions X=1 +-SQRT 5?

Vesna [10]

Answer:

D is correct. $x^2-2x-4=0$

Step-by-step explanation:

We are given the root of the equation

$x=1+\pm \sqrt{5}$

If we are given solution of the equation then find equation using formula.

If a and b are the solution of equation then equation would be (x-a)(x-b)=0

Here, $a=1+\sqrt{5} , b=1-\sqrt{5}$

Equation form would be $(x-1-\sqrt{5})(x-1+\sqrt{5})=0$

Now we simplify the above equation to get correct option.

$x^2-x+x\sqrt{5}-x+1-\sqrt{5}-x\sqrt{5}+\sqrt{5}-5=0$

$x^2-2x-4=0$

So, D is correct. $x^2-2x-4=0$

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4 years ago

Read 2 more answers

The spherical balloon is inflated at the rate of 10 m³/sec. Find the rate at which the surface area is increasing when the radiu

rjkz [21]

The balloon has a volume $V$ dependent on its radius $r$ :

$V(r)=\dfrac43\pi r^3$

Differentiating with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$

If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

$10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}$

The surface area of the balloon is

$S(r)=4\pi r^2$

and differentiating gives

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}$

so that at the moment the radius is 3 m, its area is increasing at a rate of

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}$

4 0

3 years ago