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o-na [289]
3 years ago
6

Which of the relationships you listed above will be the most helpful in figuring out the measurements of the safety fences? (2 p

oints)
Mathematics
1 answer:
defon3 years ago
3 0
Where is the options?
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Answer:  If two person had given your answer then there is a option of mark as brainliest. Click there in the option of mark as brainliest in which you like the answer most. If any one person has given answer then after some days you will a notification to mark his/her answer as brainliest. By this you can mark an answer as brainliest.

Step-by-step explanation:

4 0
2 years ago
16=2x+3x+1<br><br><br> Solve for X
stiks02 [169]
The answer is 3

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7 0
3 years ago
Read 2 more answers
Find the arc length of the partial circle.
Kay [80]

Answer:

12.42 units

Step by step explanation:

Given,

length of the radius = 2 units

Therefore, arc length of the complete circle = 2 × 3.14 × 2 units = 12.56 units

Therefore, arc length of the partial circle = 3/4 × 12.56 units

= 12.42 units

3 0
3 years ago
What is the work for 14 divided by 10.5
Softa [21]
14 divided by 10.5 is 1.3333 that goes on and on never stops. 
3 0
3 years ago
Pls help asap whats the local min value of the function below?
shutvik [7]

Answer:

<em>given function has 2 minimums</em> - \frac{9}{4}  and  \frac{9}{4}  

Step-by-step explanation:

<u><em>Step 1.</em></u> g'(x) = 4x³ - 10x

<u><em>Step 2.</em></u> Find find the critical points:

4x³ - 10x = 2x(2x² - 5) = 0

x_{1} = - \sqrt{\frac{5}{2} } , x_{2} = 0 , x_{3} = \sqrt{\frac{5}{2} }

<u><em>Step 3.</em></u> g'(x) > 0 :  - \sqrt{\frac{5}{2} } < x < 0  or  x > \sqrt{\frac{5}{2} }

g'(x) < 0 :   x < - \sqrt{\frac{5}{2} }   or   0 < x < \sqrt{\frac{5}{2} }

<u><em>Step 4.</em></u>

If x ∈ ( - ∞ , - \sqrt{\frac{5}{2} } ) , g(x) is decreasing ;

If x = - \sqrt{\frac{5}{2} } , g(x) has <em>minimum</em> value ;

If x ∈ ( - \sqrt{\frac{5}{2} } , 0 ) , g(x) is increasing ;

If x = 0 , g(x) has maximum value ;

If x ∈ ( 0 , \sqrt{\frac{5}{2} } ) , g(x) is decreasing ;

If x = \sqrt{\frac{5}{2} } , g(x) has <em>minimum</em> value ;

If x ∈ ( \sqrt{\frac{5}{2} } , ∞ ) , g(x) is increasing .

⇒ at ( - \sqrt{\frac{5}{2} } , - \frac{9}{4} ) and at ( \sqrt{\frac{5}{2} } , \frac{9}{4} ) , g(x) reaches its minimum

8 0
3 years ago
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