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3 years ago

Volume of cylinders and cones

Mathematics

1 answer:

VMariaS [17]3 years ago

8 0

Answer:

V=π r2h=cylinder

V=πr2 h/3 =cone

Step-by-step explanation:

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Answer the photo below thanks

patriot [66]

Its 1 and 2 because 5 and 4 look similar but its not as close as 1 and 2 based on the photo u put

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3 years ago

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Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.

Vesnalui [34]

Answer:

<h3>The value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$ </h3><h3>The value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$ </h3><h3>The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$ </h3><h3>The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$ </h3>

Step-by-step explanation:

Given that the Function point are $w=y^3-9x^2y$

$x=e^s$ , $y=e^t$ and s = -5, t = 10

<h3>To find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}$ using the appropriate Chain Rule : </h3>

$w=y^3-9x^2y$

Substitute the values of x and y in the above equation we get

$w=(e^t)^3-9(e^s)^2(e^t)$

$w=e^{3t}-9e^{2s}.e^t$

<h3>Now partially differentiating w with respect to s by using chain rule we have </h3>

$\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)$

$=e^{3t}-18e^{2s}.(e^t)$

$=e^{3t}-18e^{2s+t}$

<h3>Therefore the value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$ </h3>

$w=e^{3t}-9e^{2s}.e^t$

<h3>Now partially differentiating w with respect to t by using chain rule we have </h3>

$\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)$

$=3e^{3t}-9e^{2s+t}$

<h3>Therefore the value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$ </h3>

Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :

$\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}$

$=e^{3(10}-18e^{2(-5)+10}$

$=e^{30}-18e^{-10+10}$

$=e^{30}-18e^0$

$=e^{30}-18$

<h3>Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$ </h3>

$\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}$

$=3e^{3(10)}-9e^{2(-5)+10}$

$=3e^{30}-9e{-10+10}$

$=3e^{30}-9e{0}$

$=3e^{30}-9$

$\frac{\partial w}{\partial t}=3(e^{30}-3)$

<h3>Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$ </h3>

6 0

3 years ago

.A club has 18 girls and 12 boys as members. The president wants to break the club into groups, with each group containing the s

DiKsa [7]

18 girls and 12 boys 18+12= 30
Brake it up into 15

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3 years ago

How to write (-2,-1) and (2,7) in y=mx+b form

nasty-shy [4]

y = 2x + 3

the equation of a line in slope-intercept form is

y = mx + b ( m is the slope and b the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁) = (- 2, - 1) and (x₂, y₂) = (2, 7)

m = $\frac{7+1}{2+2}$ = $\frac{8}{4}$ = 2

y = 2x + b ← is the partial equation

to find b substitute either of the 2 points into the partial equation

using (2, 7 ), then

7 = 4 + b ⇒ b = 7 - 4 = 3

y = 2x + 3 ← in slope-intercept form

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4 years ago

Evaluate using suitable identity:<br>a). 103²

Simora [160]

<h3>Solution:</h3>

103^2
= (100+3)^2
= (100)^2 + 2(100)(3) + (3)^2
= 10000 + 600 + 9
= 10609

<h3>Answer:</h3>

10609

4 0

3 years ago

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