Help me please!!

Mathematics

2 answers:

statuscvo [17]2 years ago

8 0

The question is incomplete without the diagram.

Answer:

Icosahedron + Tetrahedron

Octahedron +Tetrahedron

Step-by-step explanation:

Polyhedron 1: Solid:_Icosahedron__ Number: __1___

Solid:_Tetrahedron____ Number: _20____

Polyhedron 2: Solid:___Octahedron__ Number: __1___

Solid:__Tetrahedron___ Number: __8___

dedylja [7]2 years ago

8 0

Answer:

The question is incomplete without the diagram.

Answer:

Icosahedron + Tetrahedron

Octahedron +Tetrahedron

Step-by-step explanation:

Polyhedron 1: Solid:_Icosahedron__ Number: __1___

Solid:_Tetrahedron____ Number: _20____

Polyhedron 2: Solid:___Octahedron__ Number: __1___

Solid:__Tetrahedron___ Number: __8___

Step-by-step explanation:

got it right on test

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On 5 days, lisa swam 650 meters, 675 meters, 725 meters, 800 meters, and 900 meters. What was her average distance per day?

zlopas [31]

Answer:

Her average per day was 750 meters.

Step-by-step explanation:

To find the average, add all the numbers up in the set and divide by how many numbers there are.

650 + 675 + 725 + 800 + 900 = 3750. Divide it by 5 and you get 750

6 0

2 years ago

Read 2 more answers

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$