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2 years ago

Find the surface area of the cone that has a lateral height of 4.2 and a diameter of 2.8. Use 3.14 for π.

Mathematics

1 answer:

Anvisha [2.4K]2 years ago

5 0

Answer:

here look for it there is a sol

Step-by-step explanation:

https://www.careerlauncher.com/cbse-ncert/class-9/Math/CBSE-SurfaceAreasandVolumes-NCERTSolutions.html

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Which expression should be used to estimate the answer to this story problem? A bag contained 338 beads. Three children shared t

Sedaia [141]

B. 330/3
Each kid would get 112 beads without rounding the amount of beads

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What is 41x plus 21x

Free_Kalibri [48]

The answer to that question is 62x ()//‘’::;;

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2 years ago

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Reil [10]

Is there a graph or anything given

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2 years ago

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

Now, y = vx ⇒ v = y/x

$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

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2 years ago

Wich graph matches the equition x=5

svetlana [45]

Option 1
because the x line is vertical and that’s the only one that crosses at 5

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2 years ago

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