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3 years ago

Let A = {101, 202, 303, 404, 505} and B = {101, 404}. What is A ∩ B?

Mathematics

2 answers:

Rzqust [24]3 years ago

7 0

The answers is 404 because it's in the middle

PIT_PIT [208]3 years ago

5 0

Answer:

A ∩ B = {101, 404}

Step-by-step explanation:

The given sets are A = {101, 202, 303, 404, 505} and B = {101, 404}

We have to find A ∩ B.

As we know A ∩ B is the term we use to find common terms in both.

Or intersection of two sets A and B is the set that contains all the elements of A that belongs to B, or inversely all elements of set B that belongs to A.

Now common terms of set A and set B are (101, 404).

Therefore A ∩ B = {101, 404} is the answer..

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There are 75 students performing in the holiday concert. Of the students, 30 are boys. What percent of the students are boys?

viva [34]

Answer:

Total students = 75

Boys = 30

<h3>Percentage</h3>

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Hope it helps!

8 0

2 years ago

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 ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

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).

So,

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

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$