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3 years ago

Given the parent function of y=IxI, state the type of transformation that occured to get the function below.

Mathematics

1 answer:

aksik [14]3 years ago

7 0

B. Shift down

y=|x| --------> Red
y=|x|-3 -----> Green

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Which of these equations have no solution? Check all that apply.

dlinn [17]

First one simplifies to 2x + 6 = 2x + 7 which can't be true so this one has No Solution.
Same applies to the second equation ( 5x + 15 = 5x - 15) No solution

also the last one has no solution because it simplifies to 4x + 20 = 4x + 19

6 0

3 years ago

Find a power series representation of e^x sin(x)

liberstina [14]

The Taylor series is defined by:
$f(x) = \sum \frac{f^n (a)}{n!} (x-a)^n$

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.

Derivatives can be found using product rule:
$f^n (x) = e^x g^{n-1} (x) + e^x g^n (x) \\ g^0 (x) = sin x$

Do this successively to n = 6.
$f^1 (x) = e^x (sinx +cos x) \\ f^2 (x) = e^x (2cos x) \\ f^3 (x) = e^x(2cos x - 2sin x) \\ f^4 (x) = e^x (-4 sin x) \\ f^5 (x) = e^x(-4sinx -4cos x) \\ f^6(x) = e^x(-8cos x)$

Plug in x=0 and sub into taylor series:
$e^x sin x = x+x^2 +\frac{x^3}{3} -\frac{x^5}{30}-\frac{x^6}{90}...$

If more terms are needed simply continue the recursive derivative formula and add to taylor series.

6 0

3 years ago

If six girls can finish sweeping the floor of a hall in 45mins,how long will the same work take ten girls to do

AleksandrR [38]

Answer:

15 minutes i think

Step-by-step explanation:

45÷6=7.5

meaning 7.5 minutes per girl

so 4 girls would be 30 minutes of work

so 45-30=15

4 0

3 years ago

NEED HELP ASAP! WILL MARK BRAINlIEST!

vaieri [72.5K]

Answer: 105 is your sum.

7 0

4 years ago

Read 2 more answers

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$

Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

$\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}$

Converting ÷ sign into multiplication we will take reciprocal of the second term

$=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}$

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.

5 0

3 years ago