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4 years ago

12

Liam has 9/10 gallon of paint for painting the birdhouses he sells at the

2 answers:

Masja [62]4 years ago

5 0

We know two things: The first is that Liam has 9/10 gallon of paint and the second is that each birdhouse requires 1/20 gallon of paint.
In order to calculate how many birdhouse Lian can paint with 9/10 gallon of paint we should divide 9/10 with 1/20:
(9/10) / (1/20)= 9*20/10=180/10=18
So, Liam can paint 18 birdhouses.

Archy [21]4 years ago

4 0

Liam has 9/10 of a gallon of paint, this quantity is the same as 18/20, that is,
9/10 = 18/20
9 and 10 are multiply by 2 to get 18 and 20 so that the fraction of paint available will have the same denominator as the quantity of paint that each bird house required.
The quantity of paint available is 18/20 and each house bird requires a quantity of 1/20 of gallon of paint, this means that 18 bird houses can be painted with the quantity of paint available.

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

a)Generally the concave upward interval X is mathematically given as

$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

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