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VLD [36.1K]
2 years ago
10

Find the product of the binomials. (a +1) and (a -1)

Mathematics
1 answer:
yan [13]2 years ago
6 0

Answer:

a^2-1

Step-by-step explanation:

(a+1)(a-1)

=a^2-1

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Mr.Dow lives 16 miles from his workplace he drives to and from work five days a week if his car gets 19.2 miles per gallon how m
ANEK [815]

Answer:he would require 8.33 gallons of gas

Step-by-step explanation:

Mr Dow lives 16 miles from his workplace he drives to and from work five days a week. It means that the distance that he drives everyday to and from work would be

16 × 2 = 32 miles.

Then, the total distance that he drives in a week to and from work would be

32 × 5 = 160 miles

if his car gets 19.2 miles per gallon, it means that the amount of gas that he would use in one week, commuting to and from work would be

160/19.2 = 8.33 gallons

4 0
3 years ago
If angle ABC is congruent to angle DEF, what substitution property would this equation be??? Angle DEF=180 degrees - angle ABC A
Elodia [21]

Answer:

b

Step-by-step explanation:

5 0
3 years ago
If y varies inversely as x and y=24 when x=8, find y when x is 4
olga2289 [7]
Y would equal 12 if x were to equal 4. You can think of it as a proportion.
6 0
3 years ago
What’s the answer ????
Citrus2011 [14]

Answer:

C. The graph would shift 5 units to the left

Step-by-step explanation:

If (x+5) was substituted in the place of x then it would shift 5 units to the left

Answer is C.

7 0
3 years ago
Read 2 more answers
Suppose the average tread-life of a certain brand of tire is 42,000 miles and that this mileage follows the exponential probabil
Ahat [919]

Answer: The probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

Step-by-step explanation:

The cumulative distribution function for exponential distribution is :-

P(X\leq x)=1-e^{\frac{-x}{\lambda}}, where \lambda is the mean of the distribution.

As per given , we have

Average tread-life of a certain brand of tire :  \lambda=\text{42,000 miles }

Now , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles will be :

P(X\leq 65000)=1-e^{\frac{-65000}{42000}}\\\\=1-e^{-1.54761}\\\\=1-0.212755853158\\\\=0.787244146842\approx0.7872

Hence , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

8 0
3 years ago
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