How to write one and sixth tenths as a demical

Mathematics

1 answer:

spin [16.1K]4 years ago

7 0

Answer

the answer is 1.6

Step-by-step explanation:

You might be interested in

What is the slope of a line that is perpendicular to the line whose equation is 2x + y = 4

wel

The slope of the line of the equation given is -2 (by solving for y, you get y = -2x+4).

So, in order to be perpendicular, you must take the negative reciprocal.

The slope would be -1/2.

6 0

3 years ago

Read 2 more answers

Solve the system of equations 2x-y=132x−y=13 and 6x+5y=-336x+5y=−33 by combining the equations.

andreyandreev [35.5K]

Answer: (2;-9).

Step-by-step explanation:

$\left \{ {{2x-y=13\ |*(-3)} \atop {6x+5y=-33}} \right. \ \ \ \ \left \{ {{-6x+3y=-39} \atop {6x+5y=-33}} \right. .\\Summing\ up\ these\ equations:\\8y=-72\ |:8\\y=-9.\\2x-(-9)=13\\2x+9=13\\2x=4\ |:2\\x=2.$

5 0

2 years ago

Please solve this<br><img src="https://tex.z-dn.net/?f=%20%7Cx%20%5E%7B2%7D%20%2B%203x%20-%201%20%7C%20%20%3C%203" id="TexFormul

klasskru [66]

$\bf |x^2+3x-1|\ \textless \ 3\implies 
\begin{cases}
+(x^2+3x-1)\ \textless \ 3\\\\
-(x^2+3x-1)\ \textless \ 3
\end{cases}\\\\
-------------------------------\\\\
x^2+3x-1\ \textless \ 3\implies x^2+3x-4\ \textless \ 0\implies (x+4)(x-1)\ \textless \ 0
\\\\\\
x\ \textless \ 
\begin{cases}
-4\\
1
\end{cases}\\\\
-------------------------------\\\\
-(x^2+3x-1)\ \textless \ 3\implies x^2+3x-1\boxed{\ \textgreater \ }-3
\\\\\\
x^2+3+2\ \textgreater \ 0\implies (x+2)(x+1)\ \textgreater \ 0\implies x\ \textgreater \ 
\begin{cases}
-2\\
-1
\end{cases}$

8 0

3 years ago

Which expressions are equivalent to 2-(-6+3)+4c

Artyom0805 [142]

Answer:

c. none of the above

Step-by-step explanation:

(-6+3)= -3

2--3=2+3=5

5+4c cant add them since they arent like terms

final answer 5+4c and that option isnt here

3 0

3 years ago

Plsplspslpslplspslpslsplsplsplsplsplspsl

topjm [15]

I believe the answer is 28.27, or 28.3

5 0

3 years ago