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3 years ago

What is the nearest value of d to the nearest hundredth

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

7 0

The answer is 2.05 :)

Lisa [10]3 years ago

6 0

The Answer to question is 2.05

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The graph of function f is shown.

blsea [12.9K]

Answer:

B. Both functions are increasing, but function f is increasing faster.

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3 years ago

What is the answer to this question 108=

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3 years ago

Mean for 16, 12, 13, 22

pishuonlain [190]

Answer:

mean = sum of the terms/total no. of terms

mean = (16+12+13+22)/4

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3 years ago

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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

A recipe calls for 2 1/5 cups of chopped tomatoes and 3 2/5 cups of diced turnips how many more cups of turnips did the recipe c

Zarrin [17]

The answer would be 1 1/5 more turnips.

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3 years ago

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