Solve using zero property : (x-7)(x+10) = 0

In an arithmetic series, the sum of the first 12 terms is equal to ten times the sum of the first 3 terms. If the first term is

arlik [135]

Answer:

Step-by-step explanation:

The formula for determining the sum of the first n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the arithmetic sequence.

d represents the common difference of the terms in the arithmetic sequence.

a represents the first term of the arithmetic sequence.

If a = 5, the expression for the sum of the first 12 terms is

S12 = 12/2[2 × 5 + (12 - 1)d]

S12 = 6[10 + 11d]

S12 = 60 + 66d

Also, the expression for the sum of the first 3 terms is

S3 = 3/2[2 × 5 + (3 - 1)d]

S3 = 1.5[10 + 2d]

S3 = 15 + 3d

The sum of the first 12 terms is equal to ten times the sum of the first 3 terms. Therefore,

60 + 66d = 10(15 + 3d)

60 + 66d = 150 + 30d

66d + 30d = 150 - 60

36d = 90

d = 90/36

d = 2.5

For S20,

S20 = 20/2[2 × 5 + (20 - 1)2.5]

S20 = 10[10 + 47.5)

S20 = 10 × 57.5 = 575

8 0

3 years ago

A user’s password to access a computer system consists of 3 letters followed by 2 digits. How many different passwords are possi

Vesnalui [34]

Answer:

26*26*26 = 17576 ways to select 3 letters

10*10= 100 ways to select 2 numbers

So then the total number of ways are:

$26^3 *10^2 = 1757600$ possible ways

Step-by-step explanation:

For this case we assume that we have 26 letters from A to Z and 10 numbers from 0 to 9 .

And we want to calculate the number of possible passwords possible if the password consists of 3 letters followed by 2 digits.

And for this case we can use the multiplication principle of combinatories, since we don't have any restriction about the letters of the numbers we can have repetition of letters or numbers.

For the number of possible letters:

26*26*26 = 17576 ways to select 3 letters

10*10= 100 ways to select 2 numbers

So then the total number of ways are:

$26^3 *10^2 = 1757600$ possible ways

7 0

3 years ago

Read 2 more answers

You are taking a test that you did not study for. You realize that out of the 15 questions on the test, you will need to guess o

ki77a [65]

Answer:

Step-by-step explanation:

The chance os 50%

5 0

3 years ago

Enter the first 4 terms of the sequence defined by the given rule. assume that the domain of each function is the set of whole n

OleMash [197]

The first four terms of the sequence are 8,12,16 and 20.

<h3><u>What is a Sequence?</u></h3>

A sequence is an enumerated group of items in mathematics where repetitions are permitted and order is important. Similar to a set, it has members (also called elements, or terms).
The length of the series is the number of elements (potentially infinite). In contrast to a set, the same items might appear more than once in a sequence at various points, and unlike a set, the order is important.
A sequence can be described formally as a function from natural numbers (the positions of the sequence's elements) to the items at each of those positions.
An indexed family, which is a function from an index set that may not be a set of numbers to another set of elements, can be thought of as a generalization of the idea of a sequence.

Given the function is f(n) = (2n+2)2

Now, we want first four terms, therefore, putting 1, 2, 3, 4 in the sequence we get:

f(1) = (2*1+2)2 = 8

f(2) = (2*2+2)2 = 12

f(3) = (2*3+2)2 = 16

f(4) = (2*4+2)2 = 20

Hence, The first four terms of the sequence are 8,12,16 and 20.

Know more about Sequence with the help of the given link:

brainly.com/question/21961097

#SPJ4

5 0

2 years ago

(Differential Equations) Put these equations in explicit form:

Nuetrik [128]

Answer:

1. $y= C(t*exp(-\frac{t^{2} }{2(2!)}+\frac{t^{4} }{4(4!)}-\frac{t^{6} }{6(6!)}+... ))}$

2. $t+C=ln(csc(y)-cot(y)$ $23. [tex]Assuming t as independent variable:F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\Step-by-step explanation:1. Separable variables:[tex]\frac{dy}{dt}=\frac{y*cos(t) }{t}\\ \frac{dy}{y}= \frac{cos(t) }{t}dt\\ \int {\frac{dy }{y}} \, dt=\int {\frac{cos(t) }{t}} \, dt \\ln(y)-ln(C)=ln(t)-\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+... \\y=C(t*exp(\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+...))$

2. Separable variables

\frac{dy}{sin(y)}=dt\\ \int\ \frac{1}{sin(y)}} \, dy = \int\ 1} \, dt\\t+C=ln(csc(y)-cot(y))[/tex]

3. Homogeneous D.E

Rewriting:

$dr+(\frac{1}{m} exp(m+r)+r)dt=0\\\frac{dF}{dt}=1 -> F(r,t)=t+C(r)\\\frac{dF}{dy}=0+C'(r)= \frac{1}{m} exp(m+r)+r -> C(r)=\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} \\F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\$

7 0

3 years ago