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valkas [14]
2 years ago
11

Find the sum of the first 7 terms of the following series, to the nearest integer. 150,60,24...

Mathematics
2 answers:
Kazeer [188]2 years ago
6 0

Answer:

  250

Step-by-step explanation:

The sum of the terms of a geometric series is given by the formula ...

  Sn = a1×(1 -r^n)/(1 -r)

sum of n terms for first term a1 and common ratio r.

__

<h3>series sum</h3>

The given series has first term a1 = 150, and common ratio r = 60/150 = 2/5. Putting these values into the formula gives a sum of 7 terms that is ...

  S7 = 150×(1 -(2/5)^7)/(1 -2/5) = 150((77997/78125)/(3/5))

  S7 = 150×(25999/15625) = 249.5904

Rounded to the nearest integer, the sum of the first 7 terms is 250.

Sonja [21]2 years ago
4 0

A geometric series is the collection of an unlimited number of terms with a fixed ratio between them. The sum of the first seven terms of the series is 249.

<h3>What is geometrical series?</h3>

A geometric series is the collection of an unlimited number of terms with a fixed ratio between them.

The given series is an geometric series, the details of the series are:

a₁ = 150

r = 60/150 = 0.4

n = 7

The sum of the geometric series is,

S = 150(1-0.4⁶)/(1-0.4)

S = 248.976 ≈ 249

Hence, the sum of the first seven terms of the series is 249.

Learn more about Geometrical Series:

brainly.com/question/4617980

#SPJ1

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63

Step-by-step explanation:


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3 years ago
(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2
SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(c) y_p is -4x + 7

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

(c) y'' + 2y = −8x + 14

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 16x + 7

2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

8 0
3 years ago
Julie is a cashier. She can ring up 12 customers in 9 mintesutes. At the rate, how many minutes does it take for her to ring up
oksano4ka [1.4K]

Answer:

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Step-by-step explanation:

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What happens to the diagonal of a rectangle when the rectangle is reflected across a line of symmetry? What does this suggest ab
SIZIF [17.4K]

When you reflect a diagonal over a line of symmetry, the diagonal will land perfectly on the other diagonal (and vice versa). This suggests that one diagonal is a mirror copy of the other.

Another way to put it: The vertex points of the rectangle will swap when we reflect over a line of symmetry. A diagonal is simply the opposite vertex points joined together. So this is why the diagonals swap places (because the vertices line up perfectly when you apply the reflection).

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