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frozen [14]
3 years ago
5

Let A be the area of a circle with radius r. If dr/dt=3, find dA/dt when r=2.

Mathematics
2 answers:
zlopas [31]3 years ago
5 0
Area=pir^2
take deritivive
dA/dt=2pir (dr/dt)
given dr/dt=3
dA/dt=2pi(2)(3)
dA/dt=12pi

astraxan [27]3 years ago
3 0
\bf \textit{area of a circle}=A=\pi r^2
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2r^1\cdot \cfrac{dr}{dt}\implies \cfrac{dA}{dt}=\pi \cdot 2r\cdot \cfrac{dr}{dt}\quad 
\begin{cases}
r=2\\
\frac{dr}{dt}=3
\end{cases}
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2(2)(3)
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2 years ago
Which of the following sets are subspaces of R3 ?
Ratling [72]

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The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

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\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
In the diagram, the radius of the outer circle is
JulijaS [17]

the value of x is 8 cm.

<u>Step-by-step explanation:</u>

Correct Question : In the diagram, the radius of the outer circle is 2x cm and the radius of the inside circle is 6 cm. The area of the shaded region is 220π cm2. What is the value of x? Enter your answer in the box.

We have ,

the area of a circle =  πr²

the outer circle area = \pi(2x)^2 =4\pi x^2

the inside circle area = \pi (6)^2= 36\pi

According to Question,

the outer circle area - the inside circle area = he shaded region

⇒ \pi (4x^2)-36\pi =220\pi

⇒ x^2-9 =55

⇒ x^2=64

⇒x =\sqrt{64}=8

Therefore , the value of x is 8 cm.

5 0
2 years ago
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