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hichkok12 [17]
3 years ago
6

Choose the correct solution and graph for the inequality "-x/3<=6"

Mathematics
1 answer:
lina2011 [118]3 years ago
7 0

Answer:

Step-by-step explanation:

To solve this inequality, we have to find x (or make x the subject of the inequality:

since     ⁻ˣ/₃  ≤ 6       [multiply both sides by 3]

               - x ≤ 18      [multiply both sides by -1; this flips the inequality sign]

<h3>               x ≥ - 18    [this is the solution]</h3><h3 />

To graph the inequality, draw the line x = - 18 with a solid line, and then shade the region that is x > -18.

We use a solid line because the point x = -18 is included. However, hypothetically,  if the inequality was x > -18 only, a broken line would be used to show that the line x = -18 is not included.

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mylen [45]

Answer:

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Step-by-step explanation:

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3 years ago
What is the quotient of -27.375 divided by 7.5<br><br> A)–3.65<br> B)–0.365<br> C)0.365<br> D)3.65
Anettt [7]
The answer is -3.65 because once you divide a negative with a positive the answer will still be negative.
5 0
3 years ago
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What is the domain of the function f(x) 2/5 startroot x = ? all real numbers all real numbers less than 0 all real number less t
lina2011 [118]

Answer:

\boxed{\sf all \ real \ numbers \ greater \ than \ or \ equal \ to \ 0}

Step-by-step explanation:

The domain of a function is all possible values for x.

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There are restrictions on x.

The square root of a number can be undefined if that number is less than 0.

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5 0
3 years ago
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Solve -7+11-(-3 ) what is the answer need asap for class​
finlep [7]

Answer:

=7

Step-by-step explanation:

−7+11−(−3)

=4−(−3)

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4 0
3 years ago
Considere os logaritmos log3=0,477 , log4=0,602 e / log5=0.699 logaritmos determine o valor de log5 12.
anygoal [31]

Given that :

log3=0.477 , log4=0.602  and log5=0.699

Now , as you know that

\log\text{MN}=\log M +\log N

We have to find the value of

\log_{5}12

So,\log_{5}12=\frac{\log12}{ \log5}             

                         = \frac{\log 4+\log 3}{\log 5}\\

Now Putting the values of log3, log4 and log5 in the above expression

\log_{5}12=\frac{0.477 +0.602}{0.699}

                    =1.0079/0.699

                     =1.5436..

                     =1.544 (approx)

So, the value of \log_{5}12 is 1.544(approx).



5 0
3 years ago
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