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3 years ago

(URGENT) Can someone please help me with this? Thank you! :)

Mathematics

1 answer:

Murrr4er [49]3 years ago

7 0

The remainder theorem tells you that when you do synthetic division with the 6, the remainder will be the same value as when you evaluate the function with the value of 6. The answer is 15

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50 POINTS!<br><br> Which statement best explains whether △ABC is congruent to △DEF?

aleksandr82 [10.1K]

Option B makes the most sense, because if you follow the steps, they fit perfectly into each other!

7 0

3 years ago

Read 2 more answers

Identify the sequence graphed below and the average rate of change from n = 1 to n = 3. coordinate plane showing the point 2, 8,

Alik [6]

Let $a_n$ be the n'the term of the sequence.

The first term of the sequence is $a_1$

the second $a_2$ is 8

the third term $a_3$ is 4

the fourth term $a_4$ is 2

the fifth term $a_5$ is 1

We notice that:

i) the values of the terms are powers of 2, and decreasing:

$a_2=8=2^3\\\\a_3=4=2^2\\\\a_4=2=2^1\\\\a_5=1=2^0\\\\.\\\\.$

ii) we also notice that the subscript integer of a, and the power of 2 at that term, add to 5.

thus the general term of the sequence is given by:

$a_n=2^{(5-n)}$ .

to fit the choices of the question, we can modify this formula as follows:

$a_n=2^{(5-n)}=2^{3-(n-2)}=2^3\cdot2^{-(n-2)}=8\cdot (\frac{1}{2})^{(n-2)}$

The average rate of change from n=1 to n=3 is defined as:

$\frac{a_3-a_1}{3-1}= \frac{a_3-a_1}{2}$

$a_1$ can be found using the general formula,

$a_1=2^{(5-1)}=2^{4}=16$ ,

thus the average rate of change is (4-16)/2=-12/2=-6

Answer:

$a_n=8\cdot (\frac{1}{2})^{(n-2)}$ ,

average rate of change=-6

3 0

3 years ago

The secretary in Exercises 2.121 and 3.16 was given n computer passwords and tries the passwords at random. Exactly one of the p

Vlada [557]

Answer:

There is a $\frac{(n-1)^{5}}{n^{6}}$ probability that the correct password is found on the sixth try.

Step-by-step explanation:

We have n passwords, only 1 is correct.

So,

Since a password is put back with other, at each try, we have that:

The probability that a password is correct is $\frac{1}{n}$

The probability that a password is incorrect is $\frac{n-1}{n}$ .

There are 6 tries.

We want the first five to be wrong. So each one of the first five tries has a probability of $\frac{n-1}{n}$ . So, for the first five tries, the probability of getting the desired outcome is $\frac{(n-1)^{5}}{n^{5}}$ .

We want to get it right at the sixth try. The probability of sixth try being correct is $\frac{1}{n}$ .

So, the probability that the first five tries are wrong AND the sixth is correct is:

$P = \frac{(n-1)^{5}}{n^{5}}\frac{1}{n} = \frac{(n-1)^{5}}{n^{6}}$

There is a $\frac{(n-1)^{5}}{n^{6}}$ probability that the correct password is found on the sixth try.

5 0

3 years ago

Find the time required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, co

sineoko [7]

2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

Step-by-step explanation:

The given is,

Initial investment - $5000

Future amount - $6000

Interest rate - 7.5% (compounded quarterly)

Step:1

Formula to calculate the Future amount with compound interest,

$F = P(1+\frac{r}{n} )^{nt}$ ...................................(1)

Where, F - Future amount

P - Initial amount

r - Rate of interest

n - No. of compounding in a year

t - Time period

From given,

F = $6000

P = $5000

r = 7.5%

n = 4 (compounded quarterly)

Equation (1) becomes,

$6000=5000(1+\frac{0.075}{4} )^{(t)(4)}$

$\frac{6000}{5000} =(1+0.01875)^{4t}$

$1.2 = (1.01875)^{4t}$

Take log on both sides,

$log 1.2 = 4(t) log 1.01875$

Substitute log values,

0.07918 = 4(t) (0.0080676)

= (t) (0.0322705)

$t = \frac{0.07918}{0.0322705}$

= 2.45

t ≅ 2.5 years

Result:

2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

3 0

3 years ago

How many phone numbers are possible for one area code if the first four numbers are 202-1 , in that order , and the last three n

Mkey [24]

6 phone numbers are possible for one area code if the first four numbers are 202-1

<u>Solution:</u>

Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order

We have to find how many phone numbers are possible for one area code.

The number of way “n” objects can be arranged is given as n!

Then, we have three places which changes, so we can change these 3 places in 3! ways

$n! = n \times (n - 1)!$

Hence 3! is found as follows:

$3! = 3 \times (3-1)!\\\\3! = 3 \times 2!\\\\3! = 3 \times 2 \times 1 = 6$

So, we have 6 phone numbers possible for one area code.

3 0

3 years ago