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3 years ago

Does Y= -2/3x + 3 & Y=2/3x + 2 have unlimited solutions?

Mathematics

1 answer:

notsponge [240]3 years ago

6 0

Answer:

x = 3/4

y = 2.5

Step-by-step explanation:

No there is just one.

Equate the ys

-2/3 x + 3 = 2/3 x + 2 Add 2/3 x to both sides

3 = 2/3 x + 2/3x + 2 Combine

3 = 4/3 x + 2 Subtract 2

3-2 = 4/3 x Multiply by 3

1 * 3 = 4x Divide by 4

3/4 = x

====================

y = 2/3 x + 2

y = 2/3 * 3/4 + 2

y = 6/12 + 2

y = 1/2 + 2

y = 2 1/2

y = 2.5

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Urgent please help..

Katyanochek1 [597]

Answer:

18π

Step-by-step explanation:

Lateral Surface Area = 2πrh

Plug in the numbers into the equation.

but leave the pi

6 0

3 years ago

Plz help I have a geometry exam soon!!!!! :(

Pachacha [2.7K]

Answer:

a = 93°, b = 120°, c = 150°

Step-by-step explanation:

b + 36 = 78 × 2 => b + 36 = 156 => b = 120°

87 × 2 - b = the minor arc between a and 78

=> 174 - 120 = 54

54 + b + 36 + c = 360

=> 54 + 120 + 36 + c = 360

=> c = 150°

(c + 36) ÷ 2 = a

=> (150 + 36) ÷ 2 = a

=> a = 93°

5 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

4 years ago

How do you write 133x99?

swat32

133 x 99 = 13167
100% correct

8 0

3 years ago

What is 6/8/ divided by 5/7

ziro4ka [17]

Answer:

6/8 divided by 5/7 is 1.05/ 1 1/20

8 0

3 years ago

Read 2 more answers