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nexus9112 [7]
3 years ago
6

What is the solution to 2x>6

Mathematics
1 answer:
Sliva [168]3 years ago
3 0
Divide both sides by 2 to get x>3
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Ann took
kow [346]

How do you calculate the mean (average) ?
Isn't it just

  Mean = (total points of all the tests) divided by (number of tests)  ?

Let's work with that.

Multiply each side of that formula by (number of tests), and you have

       Total points of all the tests  =  (Mean) times (number of tests).

In this problem, you know the mean, and you know the number of tests,
so you can easily calculate the total points.


3 0
4 years ago
Read 2 more answers
1, Is this triangle a right triangle? Explain.
liberstina [14]

Answer:

Yes

Step-by-step explanation:

It has the square in the corner meaning it is a right triangle, it only needs one right angle to be a right triangle. So you can say that is a shortcut.

7 0
3 years ago
In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: The length of the angle bisector of a
Aliun [14]
Check the picture below.

\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\quad cos(CAB)=\cfrac{5}{13}\implies \measuredangle CAB=cos^{-1}\left( \frac{5}{13} \right)
\\\\\\
\textit{that means that }\measuredangle hAC=\cfrac{cos^{-1}\left( \frac{5}{13} \right)}{2}\impliedby \textit{one of the halves}

now, notice, for the angle hAC, the hypotenuse is hA, and the adjacent side is CA, therefore,

\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(hAC)=\cfrac{5}{hA}\implies hA=\cfrac{5}{cos(hAC)}
\\\\\\
hA=\cfrac{5}{cos\left[ \frac{cos^{-1}\left( \frac{5}{13} \right)}{2} \right]}

make sure your calculator is in Degree mode, if you need the angle in degrees.

8 0
3 years ago
6>>>>>>>(3 3 )
alina1380 [7]
If I calculated correcty hdnnzsunssbvsbanaj?;?$;:hdmnduzvehev that is your answer
8 0
3 years ago
What is the value of b?<br> A 76 <br> B 106 <br> C 120 <br> D 224
fredd [130]
68 + a = 180
a = 180 - 68
a = 112
b = 2(112) - 104
b = 224 - 104
b = 120

answer
<span>C 120 </span>
6 0
3 years ago
Read 2 more answers
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