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3 years ago

Find the volume of a cube of edge 20 cm

Mathematics

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

8000cm^3

Step-by-step explanation:

edge(a)= 20cm

volume of cube = a^3

= 20^3

=8000 cm^3

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Anyone know the answer?

WITCHER [35]

$(f-g)(x) = f(x) - g(x) = (3x-5) - (x+3) = 2x -8$

Choice D

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3 years ago

A solution of salt and water contains 70 grams of water per 140 milliliters of the solution. if 1 mole of water weighs 16 grams,

Fynjy0 [20]

Be present in 35 milliliters of the solution. [round off your answer to two decimal places. if you answer is less than one, include the zero before the decimal point.]

6 0

3 years ago

What is the best representation of 1 and 3/4?

Rzqust [24]

the best would be 1.75

Step-by-step explanation:

3/4 turned into a decimal is .75 then you would add the 1 making it 1.75.

5 0

3 years ago

A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materi

valentina_108 [34]

Answer:

N(x) = 40 - 2x

P(x) = -2x² + 52 x - 240

maximum profit = 13

Step-by-step explanation:

given data

feeder cost = $6

average sell = 20 per week

price = $10 each

solution

we consider here price per feeder = x

and profit per feeder id here formula = x - 6

so that here

total profit will be

P (x) = ( x - 6 ) Nx

here N(x) is number of feeders sold at price = x

so formula for N (x) is here

N(x) = 20 - 2 ( x - 10 )

N(x) = 40 - 2x

so that

P(x) = (x-6) ( 40 - 2x)

P(x) = -2x² + 52 x - 240

since here

a = -2

b = 52

c = -240

a < 0

so quadratic function have maximum value of c - $\frac{b^2}{4a}$

so it will be

maximum value = -240 - $\frac{52^2}{4(-2)}$

maximum value = 98

so here maximum profit attained at

x = $\frac{-b}{2a}$

x = $\frac{-52}{2(-2)}$

x = 13

maximum profit = 13

3 0

3 years ago

(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3

Stells [14]

Answer: The required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :

$F(s)=\dfrac{16}{(s-4)^3}.$

We have the following Laplace formula :

$L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.$

Therefore, we get

$f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.$

Thus, the required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

6 0

3 years ago