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2 years ago

Please help!!!!!!!!!!!!!!!! 100pts,

Mathematics

1 answer:

Helga [31]2 years ago

3 0

QUESTION:

The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.

ANSWER:

Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–

P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.

Since there are 10 digits to choose from, we can assume that n = 10.

Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.

Now, plug these values into the formula and solve:

= 10!(10−5)!

= 10!5!

= 10⋅9⋅8⋅7⋅6

= 30240.

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The product of two consecutive integers is 420 which quadratic equation can be used to find x the lesser number

Fynjy0 [20]

Answer:

Step-by-step explanation:

If the smallest number is 'x', then the other number = x +1

x*(x+1)= 420

Use distributive property

x*x + x*1 = 420

x² + x - 420 =0 ------> this is the required quadratic equation.

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3 years ago

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IceJOKER [234]

$\frac{1}{2}$ and $\frac{1}{3}$

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3 years ago

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horrorfan [7]

Answer:
original weight of the box = 15 / 16 pounds

Explanation:
Assume that the original amount in the box is x.
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5/8 pounds represent 2/3 of the total amount (x).
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(2/3) x = 5 / 8

Now, we will solve for x as follows:
(2/3) x = 5 / 8
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Hope this helps :)

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2 years ago

Find 5 solutions to the question. Y=-1.5x+2

Black_prince [1.1K]

Answer:

☆< 《HOPE IT WILL HELP YOU》>☆

Step-by-step explanation:

Y=1.5x+2

y=1.5 (1) +2

y=1.5+2

y=3.5

y=1.5(2)+2

y=3+2

y=5

y=1.5 (3)+2

y=4.5+2

y=6.5

y=1.5 (4)+2

y=6+2

y=8

y=1.5 (5)+2

y=7.5+2

y=9.5

<h2>please mark my ans as BRAIN LIST</h2>

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3 years ago