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pogonyaev
2 years ago
15

How many blocks with dimensions of One-third times 1 times 1 can fit in a unit cube?

Mathematics
2 answers:
vodomira [7]2 years ago
8 0

Answer

Option B: 3 blocks

Step-by-step explanation

♡ ∩_∩

(„• ֊ •„)♡

┏━∪∪━━━━┓

♡ good luck 。 ♡

┗━━━━━━━┛

gtnhenbr [62]2 years ago
7 0

Answer: 3

Step-by-step explanation:

Given

The dimension of the small block is

\dfrac{1}{3}\times 1\times 1

The Volume of the cube having unit length is

V=1\times 1\times 1\\V=1\ unit^3

The number of blocks that can be in the unit cube is the division of the volume of the unit cube and block.

\Rightarrow \dfrac{1\times 1\times 1}{\dfrac{1}{3}\times 1\times 1}\\\\\Rightarrow \dfrac{3}{1}=3

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Answer:

$0.30

Step-by-step explanation:

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0.3

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4 0
3 years ago
Please help me on this problem
Orlov [11]

Answer:

The Proof with Figure, Statement and Reasons is given below.

Step-by-step explanation:

Given:

∠DAF ≅ ∠EBF,

DF ≅  FE

Prove:

Δ ADF ≅ Δ BFE

Proof:

       Statements                                 Reasons

a. ∠DAF ≅ ∠EBF       ...........................Given

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6 0
3 years ago
Determine all values of k that make each of the following a perfect square trinomial. X^2+8x+k
-Dominant- [34]

Answer:

k can either be

12

or

−

12

.

Step-by-step explanation:

Consider the equation

0=x2+4x+4

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0=(x+2)2→x=−2 and−2

. Hence, there will be two identical solutions.

The discriminant of the quadratic equation (b2−4ac) can be used to determine the number and the type of solutions. Since a quadratic equations roots are in fact its x intercepts, and a perfect square trinomial will have

2 equal, or 1

distinct solution, the vertex lies on the x axis. We can set the discriminant to 0 and solve:

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6 0
2 years ago
E
xz_007 [3.2K]

Answer:

1) n = 84

2) n = 78

3) n = 24

4) n = 16

5) n = 35

Step-by-step explanation:

To solve each proportion, we apply cross multiplication.

Question 1:

\frac{5}{12} = \frac{35}{n}

5n = 35*12

Simplifying both sides by 5

n = 7*12 = 84

Question 2:

\frac{n}{52} = \frac{180}{120}

120n = 52*180

Simplifying both sides by 20

6n = 52*9

Simplifying by 3

2n = 52*3

Simplifying by 2

n = 26*3 = 78

Question 3:

\frac{18}{n} = \frac{21}{28}

21n = 18*28

Simplifying both sides by 7

3n = 18*4

Simplifying both sides by 3

n = 6*4 = 24

Question 4:

\frac{n}{4} = \frac{24}{6}

\frac{n}{4} = 4

n = 16

Question 5:

\frac{10}{16} = \frac{n}{56}

16n = 56*10

Simplifying by 2, both sides

8n = 56*5

Simplifying by 8, both sides

n = 7*5 = 35

4 0
3 years ago
How to solve for slope? (pre algebra)
STALIN [3.7K]

Answer:

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4 0
2 years ago
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