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3 years ago

8

1)

2 answers:

borishaifa [10]3 years ago

3 0

Multiply everything in the parenthesis by -4.

-12 + 8x + 2x = 2x - 8

Combine like terms.

-12 + 10x = 2x - 8

Subtract 10x from both sides.

-12 = -8x - 8

Add 8 to both sides.

-4 = -8x

Divide both sides by -8.

x = 0.5

Hope this helps!

nasty-shy [4]3 years ago

3 0

The answer should be x=1

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3 years ago

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The product of two rational numbers is (-28/81). If one of them is (-2/3), the find the other.

KatRina [158]

Answer:

The other rational number is $\dfrac{14}{27}$ .

Step-by-step explanation:

First rational number is $\dfrac{-2}{3}$

Let other rational number is $\dfrac{x}{y}$

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According to question,

$\dfrac{x}{y}\times \dfrac{-2}{3}=\dfrac{-28}{81}$

Multiplying both sides by (-3/2) such that,

$\dfrac{x}{y}\times \dfrac{-2}{3}\times \dfrac{-3}{2}=\dfrac{-28}{81}\times \dfrac{-3}{2}\\\\\dfrac{x}{y}=\dfrac{14}{27}$

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Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

3 years ago