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Dahasolnce [82]
3 years ago
12

EXPERTS ONLY its not 0 or 128 In how many ways can you put seven marbles in different colors into four jars? Note that the jars

may be empty.
Mathematics
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:  16384

Step-by-step explanation:

Given, Total jars = 4

Total marbles = 7

Since we need to put marbles in 4 different jars, we need to choose a jar each time.

Possible choices for jars =4

Number of time we need to choose = number of marbles

So, by fundamental counting principle, we have

Total ways to put 7 marbles in 4 jars = 4\times4\times 4\times 4\times4\times 4\times 4

=4^7\\\\=16384

Hence, the required number of ways =16384

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In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.​
Karo-lina-s [1.5K]

Answer:

BC=11\ cm

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

m\angle DBC=\frac{1}{2}[arc\ DC]

substitute the values

60\°=\frac{1}{2}[arc\ DC]

120\°=arc\ DC

arc\ DC=120\°

step 2

Find the measure of arc BC

we know that

arc\ DC+arc\ BC=180\° ----> because the diameter BD divide the circle into two equal parts

120\°+arc\ BC=180\°

arc\ BC=180\°-120\°=60\°

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

m\angle BDC=\frac{1}{2}[arc\ BC]

substitute the values

m\angle BDC=\frac{1}{2}[60\°]

m\angle BDC=30\°

therefore

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step 4

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we know that

In the right triangle DBC

sin(\angle BDC)=BC/BD

BC=(BD)sin(\angle BDC)

substitute the values

BC=(22)sin(30\°)=11\ cm

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL
Mice21 [21]

Answer:

KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47

Step-by-step explanation:

Given:

KL ║ NM ,

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KN ⊥ NM  

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1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}

Also

m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ} (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

NL=45\tan 50^{\circ}

m\angle KLN=m\angle LNM=40^{\circ} (alternate interior angles)

m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

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\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

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