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3 years ago

What are the possible integer products of 8, and what are the sums of each product?

Mathematics

1 answer:

faust18 [17]3 years ago

3 0

Answer:

none of the above

Step-by-step explanation:

non of the above

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What is the value of t?

frutty [35]

Answer:

-39 2/3

Step-by-step explanation:

t - 122= 2t - 3

subtract 't' from each side to get:

-122 = 3t - 3

add 3 to each side to get:

-119 = 3t

divide each side by 3 to get:

t = -39 2/3

5 0

2 years ago

$3000 is invested at 3% annual interest for 3 years.<br> a. Simple Interest b. Compound Interest

stich3 [128]

Answer:

a. $270

b. $3,278.18

Step-by-step explanation:

Given that

The principal amount is $3,000

Annual rate of interest is 3%

And, the time period is 3 years

We need to find out the simple interest & compound interest

The following formulas should be used

a. For simple interest

= Principal × rate of interest × time period

= $3,000 × 3% × 3 years

= $270

b. For compound interest

= Principal × (1 + rate of interest)^time period

= $3,000 × (1 + 0.03)3

= $3,000 × 1.03^3

= $3,278.18

8 0

2 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

find the 20th term of an arithmetic sequence with a first term of 3 and a common difference of 5. a. 87 b. 98 c. 101 d. 130

Nady [450]

The formula for the nth term of an arithmetic sequence is:
$n_{th} \ term=a+(n-1)d$
where a is the first term and d is the common difference.
Plugging the given values into the formula, we get:
$20th\ term=3+(20-1)5=98$
The answer is b. 98.

3 0

3 years ago

What is the discriminant of 3x2 + 6x = 2?

nadezda [96]

12 times something cannot equal to 2 unless it is a negative number,is there a negative number included?

7 0

3 years ago

Read 2 more answers