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3 years ago

15

A rectangular pyramid is sliced such that the cross section is perpendicular to its base and the cross section does not intersec

t its vertex.
What is the shape of the cross section?

square

trapezoid

triangle

rectangle

2 answers:

EleoNora [17]3 years ago

5 0

Answer:

im pretty sure the answer is TRAPEZOID.

Step-by-step explanation:

o-na [289]3 years ago

5 0

The vertex, would be the tip of the pyramid. If it was sliced at the vertex, the shape would be a triangle. Since the slice doesn't intersect the vertex, the point of the pyramid would net be included, it would be as if the tip was cut off.

You would then have a trapezoid, because the top would be a straight horizontal line parallel with the bottom line.

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Answer:

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3 years ago

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Answer:

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Step-by-step explanation:

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Find the derivative of cscx/3sinx

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3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

Dale drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Dale drove hom

IrinaK [193]

Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.

Answer:

Dale live 315 miles from the mountains

Step-by-step explanation:

Let y be the speed of Dale to the mountains

Time taken by Dale to the mountains=7 hrs

Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1

Time taken by Dale back home = 5hours

Since it speed increased by 18 miles per hour back home it speed = y+18

So distance traveled home =speed × time = (y+18)5 ...... eqn 2

Since distance cover is same in both the eqn 1 and eqn 2.

Eqn 1 = eqn 2

7y = (y+18)5

7y = 5y + 90

7y - 5y = 90 (collection like terms)

2y = 90

Y = 45

Substitute for y in eqn 1 to get distance away from mountain

= 7y eqn 1

= 7×45

= 315 miles.

∴ Dale leave 315 miles from the mountains

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3 years ago