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Papessa [141]
3 years ago
8

The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance wo

uld it be possible for the cross section be a non-rectangular parallelogram?

Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0
This question has this set of answer choices:

a) when the plane cuts three faces of the cube, separating one corner from the others

b) when the plane passes through a pair of vertices that do not share a common face

c) when the plane is perpendicular to the base and intersects two adjacent vertical faces

d) when the plane makes an acute angle to the base and intersects three vertical faces

e) not enough information to answer the question

The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others

You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.

nata0808 [166]3 years ago
8 0

Answer:

It’s A

Step-by-step explanation:

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Solve: 2x+7/5 - x-3/10 = x+1/15<br>find the value of x and verify the result will RHS ​
choli [55]

Step-by-step explanation:

<h3><u>Given Question :- </u></h3>

Solve for x :-

\dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}  =  \dfrac{x + 1}{15}

\red{\large\underline{\sf{Solution-}}}

Given linear equation is

\rm :\longmapsto\: \dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{2(2x + 7) - (x - 3)}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{4x + 14 - x  +  3}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{(4x - x)  + (14 + 3)}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{3x  + 17}{10}  =  \dfrac{x + 1}{15}

On multiply by 5 on both sides,

\rm :\longmapsto\: \dfrac{3x  + 17}{2}  =  \dfrac{x + 1}{3}

On cross multiplication, we get

\rm :\longmapsto\:3(3x + 17) = 2(x + 1)

\rm :\longmapsto\:9x +51 = 2x + 2

\rm :\longmapsto\:9x  - 2x = 2 - 51

\rm :\longmapsto\:7x = - 49

\bf\implies \:x =  - 7

<h3><u>VERIFICATION</u></h3>

Consider, LHS

\red{\rm :\longmapsto\: \dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}}

On substituting the value of x, we get

\red{\rm \:  =  \:  \dfrac{2( - 7) + 7}{5} -  \dfrac{ - 7 - 3}{10}}

\red{\rm \:  =  \:  \dfrac{ - 14 + 7}{5} -  \dfrac{ - 10}{10}}

\red{\rm \:  =  \:  \dfrac{ - 7}{5}  + 1}

\red{\rm \:  =  \:  \dfrac{ - 7 + 5}{5}}

\red{\rm \:  =  \:  \dfrac{ - 2}{5}}

Consider RHS

\green{\rm :\longmapsto\:\dfrac{x + 1}{15}}

On substituting the value of x, we get

\green{\rm \:  =  \: \dfrac{ - 7 + 1}{15}}

\green{\rm \:  =  \: \dfrac{ - 6}{15}}

\green{\rm \:  =  \: \dfrac{ - 2}{5}}

\rm \implies\:LHS=RHS

HENCE, VERIFIED

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Answer:

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I don’t understand how you worded the question but x=-4 if thats what you are looking for
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