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mr_godi [17]
2 years ago
5

In the given figure, if y = 30, find (i) x (ii) z and (iii) u​

Mathematics
2 answers:
pogonyaev2 years ago
8 0

Answer:

u=30

x=150

z=150

y=30

Step-by-step explanation:

u=y [ being vertically opposite angle]

x+u=180[ being linear pair]

x+30=180

x=180-30

x=150

z=x[being vertically opposite angle]

z=150

y=u[ being voa]

velikii [3]2 years ago
7 0

Answer:

=40°

Step-by-step explanation:

Sorry that's all I know

^keep safe^

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Subtract 7x+3 from 2x-7

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3 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
A<br> 32°<br> B.<br> #<br> С<br> m2C = [?]
Dafna1 [17]

Answer:

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Step-by-step explanation:

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3 years ago
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Sati [7]
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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?
Serhud [2]

Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi }  =  \sqrt{\frac{a^{2} + b^{2} }{2} }

Squaring Both sides, we get

[\frac{C}{2\pi }]^{2}   =  [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2}  }   =  {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2}  }   =  {{a^{2} + b^{2} }

\frac{C^{2} }{2(\pi )^{2} }  = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} }  -  a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}  = b

∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

8 0
3 years ago
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