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3 years ago

Please Help what is 14.3-2^5/5

Mathematics

2 answers:

Drupady [299]3 years ago

8 0

Answer:

7.9

Step-by-step explanation:

To subtract fractions, find the LCD and then combine!

<h3>Hope this helps good luck with school!</h3>

Rashid [163]3 years ago

4 0

Answer:

The answer is 7.9

Step-by-step explanation:

14.3 - 2⁵/5

2⁵ = 32 (done by doing 2 x 2 x 2 x 2 x 2)

14.3 - 32/5

32/5 = 6.4

14.3 - 6.4 = 7.9

I hope this was helpful! If it was, please consider rating, pressing thanks, and giving my answer 'Brainliest.' Have a wonderful day! :)

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Sin (3x+19)=Cos(5x-25) What is the value of x? A) 6 B) 12 C) 18 D) 24

malfutka [58]

Answer:

Step-by-step explanation:

Using the cofunction identity

sin x = cos(90 - x)

Given

sin(3x + 19) = cos(5x - 25), then

3x + 19 = 90 - (5x - 25)

3x + 19 = 90 - 5x + 25

3x + 19 = 115 - 5x ( add 5x to both sides )

8x + 19 = 115 ( subtract 19 from both sides )

8x = 96 ( divide both sides by 8 )

x = 12 → B

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2 years ago

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matrenka [14]

Answer:

11 inch dummy

Step-by-step explanation:

4 + 7=11

4 0

3 years ago

2x - 2y = 2 and 7 = -x

Elina [12.6K]

Answer:

y = -6

Step-by-step explanation:

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3 years ago

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Kendall had $1000 in a savings account at the beginning of the semester. She has a goal to have at least $350 in the account by

USPshnik [31]

Answer:

11 weeks

Step-by-step explanation:

First we need to check what variables we have.

Beginning Balance = $1000

Goal = $350

Withdrawal = $55 per week

Now let's declare a variable as the number of weeks.

Let x = number of weeks

1000 - 55x = 350

-55x = 350-1000

-55x = -650

Then we divide both sides by -55 to find the value of x.

x = 11.81 or 11 since we're looking for how many weeks in total

Now let's see if we still have 350 if we have a total of 11 as the value of x.

1000 - 55(11) = 350

1000 - 605 = 350

395 = 350

We can see that Kendall will have $395 compared to the $350 goal.

So Kendall can withdraw $55 a week for 11 weeks to still be within her goal of having $350 in her savings account.

6 0

3 years ago

Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)

olya-2409 [2.1K]

You're looking for a solution of the form

$\displaystyle y = \sum_{n=0}^\infty a_n x^n$

Differentiating twice yields

$\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n$

$\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n$

Substitute these series into the DE:

$\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0$

$\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0$

$\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0$

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

$\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0$

which indicates that the coefficients in the series solution are governed by the recurrence,

$\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}$

Use the recurrence to get the first few coefficients:

$\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}$

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

$\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots$

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

$\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}$

Just to confirm this solution is valid: we have

y = 10x - 7eˣ ==> y (0) = 0 - 7 = -7

y' = 10 - 7eˣ ==> y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

8 0

3 years ago