All categories

4 years ago

15 POINTS

1 answer:

6 0

When a system has one answer or in other words, the graphs of the equations intersect one time, the system is called a consistent system of linear equations and therefore the equations are independent.

When a system has no solution meaning, the graphs of the equations don’t interconnect at all, the system is known to be an inconsistent system of linear equations and the equations are independent.

So therefore, the answer is yes.

You might be interested in

Which method could be used to prove they are similar?<br> 1.5<br> 3<br> 4.5<br> Jy

Veronika [31]

Answer:

Prove what is similar?????

5 0

3 years ago

Ms.green earns $908 per week for 40 hours of work. She earns 1.5 times her hourly rate for working any hours over 40 hours in a

Maurinko [17]

908÷40=22.70

22.70÷2=11.35

11.35+22.70=34.05

34.05*10=340.50

908+340.50=1,248.50

1,248.50 for 50 hours

6 0

3 years ago

Read 2 more answers

Help me!!!!!!!!!!!!!

fgiga [73]

For this case we have the following function:

$y=\frac{3}{4}x^2$

When evaluating x = 2 in the given function we have:

$y=\frac{3}{4}(2)^2$

Rewriting we have:

$y=\frac{3}{4}(4)$

$y=3$

Therefore, the pair ordered for this point is given by:

$(x, y) = (2,3)$

The ordered pair of the table for x = 2 is:

$(x, y) = (2,4)$

Therefore, the table is incorrect for x = 2

Answer:

The row in the table is incorrect for:

x=2

4 0

3 years ago

The record high January temperature in Austin, Texas, is 90 °F. The record low January is -2 °F. Find the difference between the

Setler [38]

90 - (-2) = 90 + 2 = 92....there is a difference of 92 degrees between the high and the low

3 0

4 years ago

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

3 years ago