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3 years ago

Find the circumference and the area Of the circle with diameter 9ft

Mathematics

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

Circumference is 28.26 ft

Area is 14.13 ft^2

Step-by-step explanation:

9×3.14=28.26

9/2=4.5

4.5×3.14=14.13

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Please help ASAP

yaroslaw [1]

<h3><u>SOLUTION:</u></h3>

Let the length of the room = L = 12m

width of the room = b = 8m

Height of the room = h = 4m

• Area of the four walls of the room = Perimeter of the base x Height of the room.

$= 2(l + b) \times h = 2(12 + 8) \times 4$

$= 2 \times 20 \times 4 = 160 {m}^{2}$

Cost of white washing per m² = ₹5

Hence , the total cost of white washing four walls of the room ;-

₹ ( 160 x 5 ) = ₹ 800

Area of ceiling is 12 x 8 = 96m².

Cost of white washing the ceiling.

₹ (96 x 5 ) = ₹ 480.

So , the total cost of white washing ;

₹ ( 800 + 480 ) = ₹ 1280 .

Hope this helps you :)

#Carry on learning# :) ..

8 0

2 years ago

1. four noncoplanar points<br> 2. two intersecting lines

shepuryov [24]

Whats the question here

5 0

3 years ago

There were 10 mangos 2 people had to share it equally. How many mangos does each person get?

ANTONII [103]

Each person gets 5 mangos

6 0

3 years ago

Read 2 more answers

Let C be a circle of radius 9 centered at (0,0), traversed counterclockwise. Use this curve to answer the questions below. (a) L

Katena32 [7]

a. We're looking for a scalar function $f(x,y)$ such that $\vec F(x,y)=\nabla f(x,y)$ . This means

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

Integrate both sides of the first PDE with respect to $x$ :

$\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int y\,\mathrm dx\implies f(x,y)=xy+g(y)$

Differentating both sides with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=C$

so that $\boxed{f(x,y)=xy+C}$ . A potential function exists, so the fundamental theorem does apply.

Green's theorem also applies because $C$ is a simple and smooth curve.

b. Now with (and I'm guessing as to what $\vec G$ is supposed to be)

$\vec G(x,y)=\dfrac y{x^2+y^2}\,\vec\imath-\dfrac x{x^2+y^2}\,\vec\jmath$

we want to find $g$ such that

$\dfrac{\partial g}{\partial x}=\dfrac y{x^2+y^2}$

$\dfrac{\partial g}{\partial y}=-\dfrac x{x^2+y^2}$

Same procedure as in (a): integrating the first PDE wrt $x$ gives

$g(x,y)=\tan^{-1}\dfrac xy+h(y)$

Differentiating wrt $y$ gives

$-\dfrac x{x^2+y^2}=-\dfrac x{x^2+y^2}+\dfrac{\mathrm dh}{\mathrm dy}\implies h(y)=C$

so that $\boxed{g(x,y)=\tan^{-1}\dfrac xy+C}$ , which is undefined whenever $y=0$ , and the fundamental theorem applies, and Green's theorem also applies for the same reason as in (a).

c. Same as (b) with slight changes. Again, I'm assuming the same format for $\vec H$ as I did for $\vec G$ , i.e.

$\vec H(x,y)=\dfrac x{x^2+y^2}\,\vec\imath+\dfrac y{x^2+y^2}\,\vec\jmath$

Now

$\dfrac{\partial h}{\partial x}=\dfrac x{x^2+y^2}\implies h(x,y)=\dfrac12\ln(x^2+y^2)+i(y)$

$\dfrac{\partial h}{\partial x}=\dfrac y{x^2+y^2}=\dfrac y{x^2+y^2}+\dfrac{\mathrm di}{\mathrm dy}\implies i(y)=C$

$\implies\boxed{h(x,y)=\dfrac12\ln(x^2+y^2)+C}$

which is undefined at the point (0, 0). Again, both the fundamental theorem and Green's theorem apply.

6 0

3 years ago

Your sister traveled 117 miles in 2.25 hours what is her speed as a unit rate comparing miles to 1 hour.

lord [1]

Answer:

52 miles per hour

Step-by-step explanation:

117/2.25=52

5 0

3 years ago