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3 years ago

I need help with Math please (20 points)

Mathematics

1 answer:

Afina-wow [57]3 years ago

8 0

Answer:

603

Step-by-step explanation:

The rate of change is found by

rate of change = f(b) -f(a)

------------

b-a

We need to find the rate of change for the interval 7,9 and 4,6

We know b= 9 and a = 7

f(9) -f(7)

------------

9-7

3878-1852

----------------

9-7

2026

-------

1013

We know b= 6 and a = 4

f(6) -f(4)

------------

6-4

1178-358

---------------

6-4

820

------

410

It asks how much greater, that means subtraction

1013-410

603

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3 years ago

The product of two consecutive negative even integers is 24. Find the largest integer.

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Answer:

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3 years ago

A recipe uses 5 cups of milk to make 10 servings. If the same amount of milk is used for each serving, how many servings can be

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Step-by-step explanation:

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3 years ago

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

pantera1 [17]

Answer:

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

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

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

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3 years ago