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viva [34]
3 years ago
8

What is the domain of the following parabola? (5 points) u-shaped graph that opens down with a vertex of 2, 3 x ≥ 2 x ≤ −1 y ≤ 3

All real numbers
Mathematics
2 answers:
kifflom [539]3 years ago
7 0
All real numbers would be the correct answer for this question.



Fudgin [204]3 years ago
7 0

Parabolas are U shaped graph which can open down ,up ,left or right.

Any parabola opening downwards has a maximum. The given parabola has a vertex at (2,3) or the highest point is (2,3).The domain  of any graph is the set of values x can take. For a parabola x can take any values from negative infinity to positive infinity .Any parabola can have x value as set of real numbers or we can say Domain of parabola opening down is set of real numbers.

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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500
Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking,\frac{d(w)}{dt} = 3 ft/sec

Rate at which the man is walking,\frac{d(m)}{dt} = 2 ft/sec

Collective rate of both, \frac{d(m+w)}{dt} = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min  = 20 min

Now,

Distance traveled by man and woman are m and w ft respectively.

⇒ m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft

⇒ w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\  ft

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate \frac{dh}{dt} .

Lets calculate h.

Applying Pythagoras formula.

⇒ h^2=(m+w)^2+500^2  

⇒ h=\sqrt{(2400+2700)^2+500^2} = 5124.45

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ h^2=(m+w)^2+500^2

⇒ 2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt}  + \frac{d(500)}{dt}

⇒ \frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}                         ...as \frac{d(500)}{dt}= 0

⇒ Plugging the values.

⇒ \frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}                       ...as \frac{d(m+w)}{dt} = 5 ft/sec

⇒ \frac{d(h)}{dt} =4.9761  ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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