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3 years ago

Find the area of the circle r = 3 ft

Mathematics

1 answer:

pav-90 [236]3 years ago

5 0

Answer:

9π ft² or approximately 28.26 ft²

Step-by-step explanation:

The area of a circle is A=πr².

Plug 3 in for the radius.

A=π(3²)

A=9π ft²,

or using 3.14 for pi,

Approximately 28.26 ft².

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labwork [276]

Domain is the numbers you can use
range is the result of inputing the domain

an interesting fact is that the inverse of a function switches the domain and range

basically
the domain of f(x) becomes the range of f^-1(x)
the range of f(x) becomes the domain of f^-1(x)
so just find the domain and range of f(x)

$f(x)=3x- \frac{1}{2}$
there are no restrictions
all real numbers can be used
all real numbers can result

so the answer is domain and range for both is all real numbers

D is answer

7 0

3 years ago

(6 × 10^4 + 3 × 10^4)(3 × 10^2) What is the value of the expression in scientific notation?

LekaFEV [45]

Answer:

2.7x10^7 or 2.70000000 I think

Step-by-step explanation:

6 0

3 years ago

Set A contains 35 elements and set B contains 22 elements. If there are 40 elements in (A ∪ B) then how many elements are in (A

Leona [35]

Answer:

Step-by-step explanation:

The computation of the number of elements are in (A ∩ B) is shown below;

Given that

Set A contains 35 elements

And, set B contains 22 elements

Now if there are 40 elements in (A ∪ B)

So, the number of elements are in (A ∩ B) is

= 35 + 22 - 40

= 17

5 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

Simplify the expression 17y-2x+3(3x-y)+2x

Marat540 [252]

Answer:

9x+14y

Step-by-step explanation:

First Multiply the number outside of the brackets with the numbers inside the brackets. Ex:3 x 3x and 3 x -y

17y-2x+3(3x-y)+2x

That should get you to this.

17y-2x+9x-3y+2x

Next combine like terms.

14y+9x

Hope this helps!

5 0

4 years ago