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3 years ago

Find the sum of the first 10 terms of the following series, to the nearest integer.

Mathematics

1 answer:

Alona [7]3 years ago

4 0

Answer:

Step-by-step explanation:

you can see this is a geometric progression which a ratio of ${2\over{5}}$ . the sum of the firs n term is:

$S=a_1\frac{(1-r^n)}{(1-r)}=175\frac{1-(\frac{2}{5})^{10}}{1-\frac{2}{5}}=291.6360832$

the nearest integer would be:

292

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Credit unions are not subject to federal regulations. A. True B. False

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Answer:

The correct answer is (FALSE)

Step-by-step explanation:

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3 years ago

Two data sets have the same mean,

nirvana33 [79]

For Data Set B, we see that the data is more varied. The absolute deviations are 4, 3, 2, 5. The average of these absolute deviations is 3.5. MAD_B = (4+3+2+5)/4 =3.5 M ADB

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2 years ago

An online ticket seller charges $44 for each ticket to a concert, plus a fixed handling fee of $12. Define the unknown variables

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The unknown variables are how many tickets are being bought and the total.

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3 years ago

How can a percent be used to describe a change in quantity?

Anestetic [448]

Answer:

remember,a percent means "per 100".You can describe the change as a fraction comparing the amount of change to the original and then use a proportion to rewrite your problem as a percentage out of 100.

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2 years ago

Help. <br>Please its urgent show workings.<br>

laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

$\frac{dy}{dx}$ = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × $\frac{d}{dx}$ (x + 4) ← chain rule

= 2(x + 4) × 1

= 2(x + 4)

Then

$\frac{dy}{dx}$ = (2x - 1). 2(x + 4) + (x + 4)². 2

= 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

= 2(x + 4) (2x - 1 + x + 4)

= 2(x + 4)(3x + 3) ← factor out 3

= 6(x + 4)(x + 1)

--------------------------------------------------------------------------

(b)

y = x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × $\frac{d}{dx}$ (x² - 1) ← chain rule

= 3(x² - 1)² × 2x

= 6x(x² - 1)²

Then

$\frac{dy}{dx}$ = x. 6x(x² - 1)² + (x² - 1)³. 1

= 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

= (x² - 1)² (6x² + x² - 1)

= (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

$\frac{dy}{dx}$ = (x² - 1). 3x² + (x³ + 1), 2x

= 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

= x[3x(x² - 1) + 2(x³ + 1) ]

= x(3x³ - 3x + 2x³ + 2)

= x(5x³ - 3x + 2) ← distribute

= 5 $x^{4}$ - 3x² + 2x

--------------------------------------------------------------------

(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × $\frac{d}{dx}$ (x² + 4) ← chain rule

= 2(x² + 4) × 2x

= 4x(x² + 4)

Then

$\frac{dy}{dx}$ = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

= 12 $x^{4}$ (x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)

= 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

= 3x²(x² + 4)(4x² + 3x² + 12)

= 3x²(x² + 4)(7x² + 12)

5 0

3 years ago