Answer:
y + 1 + y + 2 + y + 3 =3y+6
Step-by-step explanation:
study hard:)
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.
The perimeters of both are equal.
The side of the square is 12.
Therefore, its area equals to : 12² = 144
The rectangle base is 19.
Because it has the same perimeter as the square's, so rectangle perimeter is : 19 + 19 + side + side = 12 + 12 + 12 + 12
= 38 + 2side = 48
= 2side = 48 - 38
= side of rectangle = 5
Therefore, its area is 19 x 5 = 95
If you subtract it from the area of the square, you will get : 144 - 95 = 49.
So the answer is : the area of the square is 49 units largee than the area of the square (C)
Answer:
x=-2
y= 8
Step-by-step explanation:
Given data
10x – 2y = -36-----------1
7x – 2y = -30--------------2
-Subtract to eliminate y
3x-0= -6
3x= -6
x= -6/3
x= -2
Put x= -2 in
7(-2)– 2y = -30
-14-2y=-30
-14+30=2y
16=2y
y= 16/2
y= 8
Answer:
average atomic mass = 207.2172085 amu
Explanation:
To get the average atomic mass of an element using the abundance of its isotopes, all you have to do is multiply each isotope by its percentage of abundance and then sum up all the products.
For the question, we have:
203.97304 amu has an abundance of <span>1.390% (0.0139)
</span>205.97447 amu has an abundance of 24.11% (0.2411)
206.97590 amu has an abundance of 22.09% (0.2209)
207.97665 amu has an abundance of 52.41 % (0.5241)
The average atomic mass can be calculated as follows:
average atomic mass = 203.97304(0.0139) + 205.97447(0.2411)
+ 206.97590(0.2209) + 207.97665(0.5241)
average atomic mass = 207.2172085 amu
Hope this helps :)
