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2 years ago

Find the vector, not with determinants, but by using properties of cross products. k × (i − 3j)

1 answer:

3 0

$\mathbf k\times(\mathbf i-3\mathbf j)=(\mathbf k\times\mathbf i)-3(\mathbf k\times\mathbf j)$
$=\mathbf j-3(-\mathbf i)$
$=3\mathbf i+\mathbf j$

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Define Following:consistent equations

mario62 [17]

Answer:

1. Consistent equations

x + y = 3

x + 2·y = 5

2. Dependent equations

3·x + 2·y = 6

6·x + 4·y = 12

3. Equivalent equations

9·x - 12·y = 6

3·x - 4·y = 2

4. Inconsistent equations

x + 2 = 4 and x + 2 = 6

5. Independent equations

y = -8·x + 4

8·x + 4·y = 0

6. No solution

4 = 2

7. One solution

3·x + 5 = 11

x = 2

Step-by-step explanation:

1. Consistent equations

A consistent equation is one that has a solution, that is there exist a complete set of solution of the unknown values that resolves all the equations in the system.

x + y = 3

x + 2·y = 5

2. Dependent equations

A dependent system of equations consist of the equation of a line presented in two alternate forms, leading to the existence of an infinite number of solutions.

3·x + 2·y = 6

6·x + 4·y = 12

3. Equivalent equations

These are equations with the same roots or solution

e.g. 9·x - 12·y = 6

3·x - 4·y = 2

4. Inconsistent equations

Inconsistent equations are equations that are not solvable based on the provided set of values in the equations

e.g. x + 2 = 4 and x + 2 = 6

5. Independent equations

An independent equation is an equation within a system of equation, that is not derivable based on the other equations

y = -8·x + 4

8·x + 4·y = 0

6. No solution

No solution indicates that the solution is not in existence

Example, 4 = 2

7. One solution

This is an equation that has exactly one solution

Example 3·x + 5 = 11

x = 2

3 0

3 years ago

Which describes money that is received from a sale of goods or services?

dalvyx [7]

The answer is Revenue

6 0

2 years ago

Heather works part time at a movie theater where she earns $7.00

Nastasia [14]

Answer:

d. 1 grid equals 1 hour

Step-by-step explanation:

When plotting research data, X-axis(or horizontal axis) usually used for independent variable and Y-axis is used for the dependent variable. In this case, Heather wants to know how much earning on different numbers of hours. The dependent variable is the earning and the independent variable is the hours, so you put hours on the horizontal axis.

You want to make a 10x10 grid of data and the hours ranged between 1-10. If you plot them equally, approximate scale will be: (10h-1h)/(10)= 0.9h/grid

The closest option is 1 hour per grid. It will provide the best visualization since it won't stretch or minimize the data too much.

8 0

2 years ago

Solve for x 5x+7=9x-25

Svetlanka [38]

Subtract 7 both sides
Subtract 9x both side
They cancel out
At the end you divide it
I don’t have access to paper right now sorry

6 0

3 years ago

Read 2 more answers

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

2 years ago