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Alenkasestr [34]
3 years ago
15

Solve and explain please.

Mathematics
1 answer:
deff fn [24]3 years ago
4 0
Are you still doing it.?
You might be interested in
Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132
Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

The HCF is the product of the highest factors

So, the HCF is:

HCF = 2 * 3 * 11

HCF = 66

<h3>How to determine the LCM</h3>

In (a), we have:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

So, the LCM is:

LCM = 2^3 * 3 * 7 * 11

LCM  = 1848

Hence, the HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

4 0
2 years ago
Your friend claims that it is possible to draw a right triangle where the cosine of either angle (theta) is exactly the same val
avanturin [10]

Answer:

B. No

Cos \ \theta\neq Cos (90-\theta)\textdegree

Step-by-step explanation:

-A right angle triangle has two complimentary acute angles and one right angle.

-\theta is usually one of the acute angles and is equivalent to 90º minus it's complimentary acute angle.

-Complimentary angles add up to 90º.

#For complimentary angles:

Sin \ \theta=Cos \ (90-\theta)\textdegree\\\\Cos \ \theta=Sin(90-\theta)\textdegree\\\\\therefore Cos \ \theta\neq Cos (90-\theta)\textdegree

The two acute angles cannot have the same Cosine value.

Hence, she's not correct.

5 0
3 years ago
Students were selected at random in a local high school to see if there was an association between gender and
Assoli18 [71]
<h3>Answer:  53%</h3>

===========================================================

Explanation:

The question asks "What percent of females participate in extracurricular activities?" This means we only focus on the second column. There are 36 women, and of this total, 19 are in extracurricular activities.

Dividing the two values leads to approximately 19/36 = 0.52777 which rounds to 0.53

Then we move the decimal to the right two spots to get 53%

Roughly 53% of the female students participate in extracurricular activities.

5 0
3 years ago
A crayon company sells a pack of wax crayons for 5$. There are 10 crayons in each pack. What is the unit price of the crayons
Alchen [17]

10 crayons = $5

1 crayon = $5 ÷ 10 = $0.50


Answer: Unit Price = $0.50/crayon

5 0
3 years ago
Read 2 more answers
A ________ exists between two variables when the values of one variable are somehow associated with the values of the other vari
motikmotik
Correlation would be the correct answer
8 0
3 years ago
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