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Natasha2012 [34]
3 years ago
10

What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
6 0

Answer:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

Hence, ption B is true.

Step-by-step explanation:

Given the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

solving the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

as

\sqrt{x^2y^3}=xy\sqrt{y}

2\sqrt{x^3y^4}=2xy^2\sqrt{x}

so the expression becomes

\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}

Group like terms

                                        =2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}

Add similar elements

                                        =2xy^2\sqrt{x}+2xy\sqrt{y}

Therefore, we conclude that:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

Hence, option B is true.

mylen [45]3 years ago
5 0

Answer:

b

Step-by-step explanation:

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The equation cos−1 = x can be used to determine the measure of angle BAC.
Sergio [31]

we know that

in the right triangle ABC

the value of cosine of angle x is equal to

cos(x)=\frac{AC}{AB}

we have

AC=3.4\ cm\\AB=10\ cm

substitute the values

cos(x)=\frac{3.4}{10}=0.34

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x=cos(x)^{-1} =arc\ cos(0.34)=70.12\°

Round to the nearest whole degree

x=70\°

therefore

<u>the answer is</u>

70\°


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Step-by-step explanation:

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Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
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<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
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In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
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<span>d+1=−6</span> => <span>d=−7</span>
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Question is on the photo. Thanks!
Elanso [62]

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Step-by-step explanation:

f(x) = x + 2

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