Evaluate 2(L+W)<br> for L=10 and W= 2

What are 25 times by the same number then divide it by sixty

larisa86 [58]

25(25) = 625

625/60 ≈ 10.4166

Or when rounded to the nearest hundredth, 10.42

6 0

3 years ago

The 1,840 rock concert tickets sold were twice the amount of jazz concert tickets sold

Yuki888 [10]

So there where two times the amount of rock tickets sold than the jazz concert. If that is it then you divide the 1840 by 2 getting 920, dividing 1÷2 you get zero remainder 1, 18÷2= 9, 4÷2= 2, and 0÷2= 0. 920. Hope i helped :)

8 0

3 years ago

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Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

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On a certain hot Summer's day 388 people use the public swimming pool. The daily prices are $1.75 for children and $2 for adults

Alecsey [184]

Answer:

it could be 372 adults with one kid or 426 kids with no adults but that isn't logical

Step-by-step explanation:

i don't really know how to explain it sorry

6 0

3 years ago

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0

3 years ago

Evaluate ​2(L+​W) for L=10 and W= 2

Evaluate 2(L+W) for L=10 and W= 2