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3 years ago

I need help on this plz i don’t understand the question

Mathematics

1 answer:

AnnZ [28]3 years ago

3 0

We know that the area of a circle in terms of π will be πr². However the area with respect to the diameter will be a different story. The first step here is to find a function relating the area and diameter of any circle --- ( 1 )

For any circle the diameter is 2 times the radius,

d = 2r

Therefore r = d / 2, which gives us the following formula through substitution.

A = π(d / 2)² = πd² / 4

<u>Hence the area of a circle as the function of it's diameter is A = πd² / 4. You can also say f(d) = πd² / 4.</u>

Now we can substitute " d " as 4, solving for the area ( A ) or f(4) --- ( 2 )

f(4) = π(4)² / 4 = 16π / 4 = 4π - <u>This makes the area of circle present with a diameter of 4 inches, 4π.</u>

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3 years ago

I need help ASAP please

Solnce55 [7]

Answer:

x = 6

Step-by-step explanation:

The area of a triangle is equal to the base times the height divided by two. In this case, the base is 6 (or x, it doesn't really matter) and the height is x (or 6, it doesn't matter). That means that the area of this triangle is equal to 6 * x divided by 2, which is also equal to 18. we can say that 6x/2 = 18, meaning that 6x = 36, so x = 6.

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2 years ago

Factor the polynomial by it’s greatest common monomial factor. <br> 12k^3 - 30k^4

madam [21]

Answer:

6k^3(2 - 5k)

Step-by-step explanation:

12k^3 - 30k^4

Factor out 6k^3

6k^3(2 - 5k)

5 0

3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

A manufacturer produces bolts of a fabric with a fixed width. A quantity q of this fabric (measured in yards) that is sold is a

BigorU [14]

Answer:

$R'(20) = 2000$

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

$q=f(p)$

Total revenue earned =

$R(p)=pf(p)$

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

$R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)$

Putting the values, we get,

$R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000$

7 0

4 years ago