What is the formula for the accounting equation

Segments UV and WZ are parallel with line ST intersecting both at points Q and R respectively

Zanzabum

The <em><u>correct answer</u></em> is:

Corresponding (for both blanks)

Explanation:

Corresponding angles are angles that are in the same relative position to both the parallel line they are on and the transversal. In this case, SQU and WRS are both above the parallel line they are on and on the same side of the transversal; this makes them corresponding angles. This also means it is the corresponding angles postulate.

3 0

3 years ago

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What's x and y for x=1+y and 2y+x=10

Sav [38]

Substitute the value of x in equation 2. Then you'll get y=3. Then again substitute the value of y in x=1+y, you'll get x=4

8 0

3 years ago

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If two matrices, A and B, are equal, which of the

alexandr402 [8]

If A and B are equal:

Matrix A must be a diagonal matrix: FALSE.

We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:

$A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]$

Both matrices must be square: FALSE.

We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works

Both matrices must be the same size: TRUE

If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.

For any value of i, j; aij = bij: TRUE

Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.

8 0

3 years ago

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A square has a side length equal to 5x-3y inches. What is the area of this square

erastovalidia [21]

Answer:

D.

Step-by-step explanation:

The area is the product of 2 adjacent sides:

Area = (5x - 3y)(5x - 3y) as the sides of a square are all equal.

= 5x(5x - 3y) - 3y(5x - 3y)

= 25x^2 - 15xy - 15xy + 9y^2

= 25x^2 - 30xy + 9y^2.

5 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago