Answer:
perimeter: 12x-6
area: 6x-10
Step-by-step explanation:
PERIMETER:
6x-5(2)+2(2)
12x-10+4
12x-6
AREA:
6x-5•2
6x-10
11.333333 (three repeated, don't know how to type that
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.
Answer:
can be factored out as: 
Step-by-step explanation:
Recall the formula for the perfect square of a binomial :

Now, let's try to identify the values of
and
in the given trinomial.
Notice that the first term and the last term are perfect squares:

so, we can investigate what the middle term would be considering our
, and
:

Therefore, the calculated middle term agrees with the given middle term, so we can conclude that this trinomial is the perfect square of the binomial:

Answer:
m∠ABD = 88º
m∠CBD = 23º
Step-by-step explanation:
(-10x + 58) + (6x + 41) = 111
Combine like terms
-4x + 99 = 111
Subtract 99 from both sides
-4x = 12
Divide both sides by -4
x = -3
------------------------
m∠ABD = -10x + 58
m∠ABD = -10(-3) + 58
m∠ABD = = 30 + 58
m∠ABD = 88º
m∠CBD = 6x + 41
m∠CBD = 6(-3) + 41
m∠CBD = -18 + 41
m∠CBD = 23º