What is the hcm of 18 and 45

Mathematics

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

Step-by-step explanation:

90 Or, we can say that it is a method of breaking down the given integer into its prime factors. Therefore, the HCF of 18 and 45 is 9. Therefore, the HCF of 18 and 45 is 90.

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levacccp [35]

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3 years ago

Miranda made 6.88 pounds of trail mix. She used 4.96 pounds of granola for the mix. Miranda wrote the equation w + 4.96 = 6.88 t

joja [24]

Step 1: to solve this equation first subtract 4.9 from both sides.

w + 4.9 - 4.9 = 6.88 - 4.9

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3 years ago

Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x

svetlana [45]

Hello,

$sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

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= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\

$
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$2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

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= \dfrac{1}{sin^2x*cos^2x}} 
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8 0

4 years ago

Read 2 more answers

principle is used to find the number of strings of eight uppercase English letters that start or end with the letters BO (in tha

yKpoI14uk [10]

There are 6 other positions in the string, each with 26 choices. So if you fix BO as the first two letters, there are $26^6$ possible strings that you can make.

If BO is at the end of the string, you still have $26^6$ possible strings.

Together, then, you have $2\times26^6$ possible strings.