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Solve the inequalityCould you please be fast I only have q hour to finish and I still have a bunch of questions to finish

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Answer 1

Answer:

Let's solve the inequality

$\begin{gathered} -x\leq15-2x \n 2x-x\leq15 \n x\leq15 \end{gathered}$

Therefore the solution of the inequality is the set:

$(-\infty,15\rbrack$

$w \leq \: 26$

Step-by-step explanation:

$(9w - 4)/(5) \leq (7w + 2)/(4)$

First of all cross multiply

That's

$4(9w - 4) \leq5(7w + 2)$

Expand the terms

$36w - 16 \leq \: 35w + 10$

Add 16 to both sides

That's

$36w + 16 - 16 \leq \: 35w + 10 + 16 \n 36w \leq35w + 26$

Subtract 35w from both sides

We have

$36w - 35w \leq35w - 35w + 26 \n$

We have the final answer as

$w \leq \: 26$

Hope this helps you

I enter the garden. There are 34 people. You hug 30. How many people are in the garden.

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There are 34 people in the garden prior to the time I joined them.

So when I entered the garden, the total number of people in the garden will be:

34 + 1 = 35 people

Let assume, you as the reader was part of the people in the garden and you hugged 30 people. That doesn't add up or increase the number of people in the garden.

At the end of the day, the number of people you did not hug is:

35 - 30 - 1(you)

= 35 - 31

= 4 people

Therefore, we can conclude that there are 35 peoplepresent in the garden.

Learn more about word problems here:

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Answer:

Step-by-step explanation:

Ryan and Taylor are both saving money to buy new video game

Answers

Hope they get enough saved.

Find the missing value for each of the following 3 questions 1. 5% of 60 is _____

2. 40% of 500 is _____

3. 75% of 60 is ______

Answers

Answer:

1. 3

2. 200

3. 45

Step-by-step explanation:

For the characteristic polynomialp(s) =s5+ 2s4+ 24s3+ 48s2−25s−50(a) Use the Routh-Hurwitz Criterion to determine the number of roots ofp(s) in the right-half plane, in the left-half plane, and on thejω-axis.(b) Use Matlab to determine the roots ofp(s), and verify your results in part 2a.

Answers

Answer:

1 root in the right half-plane
1 conjugate pair on the imaginary axis
2 roots in the left half-plane

Step-by-step explanation:

Without using the Routh-Hurwitz criterion at all, you know there is one positive real root. Descartes' rule of signs tells you the number of positive real roots is equal to the number of sign changes in the coefficients (perhaps less a multiple of 2). There is one sign change in + + + + - - , so there is one positive real root.

_____

(a) The Routh array starts as two rows of the polynomial's coefficients, alternate coefficients on each row. For this odd-degree polynomial, the number of coefficients is even, so no zero-padding is necessary at the right end of the second row. That is, we start with ...

$\begin{array}{cccc}s^5&1&24&-25\ns^4&2&48&-50\end{array}$

The next row is formed from combinations of coefficients in the two rows above. The computation is similar to that of a determinant. By matching the numbers to those in the array, you can see the pattern of the computation.

The next row values are ...

$\begin{array}{ccc}s^3&((2)(24)-(1)(48))/(2)&((2)(-25)-(1)(-50))/(2)\end{array}$

Simplifying, we find this row to be ...

$\begin{array}{ccc}s^3&0&0\end{array}$

The zero row is a special case that requires we proceed as follows. The row above (identified with s⁴) represents an "auxiliary polynomial":

$2s^4 +48s^2 -50$

To continue the process, we replace the zero row by the coefficients of the derivative of this auxiliary polynomial. Proceeding as before, the array now becomes ...

$\begin{array}{cccc}s^5&1&24&-25\ns^4&2&48&-50\ns^3&8&96\ns^2&24&-50\ns^1&112(2)/(3)&0\ns^0&-50\end{array}$

The number of sign changes in the first column (1) tells the number of roots in the right half-plane. The auxiliary polynomial will give us the remaining two pairs of roots:

$2s^4+48s^2-50=0\n\n2(s^2+25)(s^2-1)=0\n\ns=\pm 5i,\ s=\pm 1$

So, we have determined there to be ...

1 root in the right half-plane
2 roots on the jω axis
2 roots in the left half-plane

(b) The original polynomial can be factored as ...

p(s) = (s +2)(s² +25)(s +1)(s -1)

p(s) = (s +2)(s +1)(s -5i)(s +5i)(s -1)

This verifies our result from part (a).

_____

Additional comments

Any row can be multiplied by a convenient factor to simplify the arithmetic. Here, it would be convenient to divide the second row by 2 and the third row by 8.

A zero element (not row) in the first column is replaced by "epsilon" (a small positive number) and the rest of the arithmetic is continued as normal. That row is not counted (it is ignored) when counting sign changes in the first column.

On January 1, 2021, White Water issues $410,000 of 7% bonds, due in 10 years, with interest payable semiannually on June 30 and December 31 each year.Assuming the market interest rate on the issue date is 8%, the bonds will issue at $382,141.

Record the bond issue on January 1, 2021, and the first two semiannual interest payments on June 30, 2021, and December 31, 2021. (If no entry is required for a particular transaction/event, select "No Journal Entry Required" in the first account field. Round your final answers to the nearest whole dollar.)

Answers

The appropriate journal entries to record the bond issue on January 1, 2021, and the first two semiannual interest payments on June 30, 2021, and December 31, 2021 are:

White Water journal entries

1-Jan-21

Debit Cash $382,141

Credit Discount on Bonds Payable $27,859

($410,000-$382,141)

Credit Bonds payable $ 410,000

30-Jun

Debit Interest Expenses $ 15,286

($382,141 x 8%/2)

Debit Discount on Bonds Payable $736

Credit Cash $14,350

($410,000 x 7%/2)

31-Dec

Debit Interest Expenses $15,315.08

[($382,141 + 736) x 8%/2]

Credit Discount on Bonds Payable $965.08

($15,315.08-$14,350)

Credit Cash $14,350

($410,000 x 7%/2)

Learn more here:

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Answer:

Step-by-step explanation:

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Solve the inequalityCould you please be fast I only have q hour to finish and I still have a bunch of questions to finish

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