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Andru [333]
4 years ago
15

An image of a rectangular pyramid is shown below: A right rectangular pyramid is shown. Part A: A cross section of the rectangul

ar pyramid is cut with a plane parallel to the base. What is the name of the shape created by the cross section? Explain your answer. (5 points) Part B: If a cross section of the rectangular pyramid is cut perpendicular to the base, passing through the top vertex, what would be the shape of the resulting cross section? Explain your answer. (5 points)

Mathematics
1 answer:
valentina_108 [34]4 years ago
5 0

Answer:

Part a) Rectangle

Part b) Triangle

Step-by-step explanation:

<u><em>The picture of the question in the attached figure N 1</em></u>

Part A) A cross section of the rectangular pyramid is cut with a plane parallel to the base. What is the name of the shape created by the cross section?

we know that

When a geometric plane slices any right pyramid so that the cut is parallel to the plane of the base, the cross section will have the same shape (but not the same size) as the base, So, in the case of a right rectangular pyramid, the cross section is a rectangle

Part b)  If a cross section of the rectangular pyramid is cut perpendicular to the base, passing through the top vertex, what would be the shape of the resulting cross section?

we know that

Cross sections perpendicular to the base and through the vertex will be triangles

see  the attached figure N 2 to better understand the problem

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Step-by-step explanation:

Hola, para resolver este problema simplemente hay que sumar las cantidades de cada fruta:

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Matemáticamente hablando:

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A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in
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Answer: The concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

                               Compound X             Compound Y

272                               16400                            3870

327                                3990                             6420

So, our equations becomes

16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191

By solving these two equations, we get that

C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}

Hence, the concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

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