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3 years ago

6

A combination lock requires 4 selections of numbers, each from 1 to 20. • how many different combinations are there? • suppose t

hat the lock is constructed in such a way that no number maybe used twice. how many different combinations are possible?

1 answer:

slamgirl [31]3 years ago

6 0

NUmber of combination = 20 x 19 x 18 x 17 = 116 280

Answer: 116 280

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If a sentence is defined recursively by f(0)=3 and f(n+1) = f(n) +5 for n ? 0, then f(2) is A) 2 B) 3 C) 5 D) 8

VladimirAG [237]

The recursive formula $f(n+1)=f(n)+5$ means that the next term in the sequence is 5 more than the previous one.

So, we know that we start from $f(0)=3$ , which means that the next term is

$f(1) = f(0)+5 = 3+5 = 8$

Similarly,

$f(2) = f(1)+5 = 8+5 = 13$

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3 years ago

Given that h(x) = 5-2x, find the value(s) of x when h(x) =3

arsen [322]

Answer:

x=1

Step-by-step explanation:

We have,

h(x) = 5-2x

h(x)=3

Now,

h(x)=5-2x=3

or,5-2x=3

or,5-3=2x

or,2/2=x

Therefore, x=1 when h(x)=3

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2 years ago

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Vitek1552 [10]

-24y = -384, y would equal 16.

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3 years ago

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The HCF and LCM of two numbers x and 126 are 24 and 840 respectively. Find the value of x.

Zigmanuir [339]

Answer:

x=160

Step-by-step explanation:

x×126=24×840

$x=\frac{24 \times840}{126} =160$

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3 years ago

If x=5^a, find the value of 5x. How?

Yakvenalex [24]

Answer:

$5x = 5^{(a+1)}$

Step-by-step explanation:

Here is a problem with the concept of properties of the exponent.

We are given $x = 5^{a}$ and we have to find the value of 5x.

Now, $5x = 5 \times 5^{a}$ {Since we are given that $x = 5^{a}$ }

⇒ $5x = 5^{1} \times 5^{a}$ {Since we can write $x = x^{1}$ }

⇒ $5x = 5^{(a+1)}$ {Since we know the formula of exponent as $x^{m} \times x^{n} = x^{(m + n)}$ }

(Answer)

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3 years ago