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Mrrafil [7]
3 years ago
9

SURDS

Mathematics
1 answer:
aev [14]3 years ago
4 0
1) √3 √7 = √21
2) √5 √245 = √5 √5 * 49 = √5 * 7√5 = 7 √5 * 5 = 7 √25 = 7 * 5 = 35
3) √77 ÷ √11 = as is. can't be simplified.
4) (√59)² = 59 ; the square root was cancelled by squared.
5) 3√6 x 8√7 = 3 * 8 √6 * 7 = 24 √42
6) 5√3 x 6 √3 = 5 * 6 √3 * 3 = 30 √9 = 30 * 3 = 90
7) 40√30 ÷ 5√3 = (40 / 5) * (√30 /√3) = 8 * ((√3 *10) / √3) = 8 √10
8) (6√5)² = 6² * √5² = 36 * 5 = 180
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3 years ago
Need help with this<br>​
AleksandrR [38]

Step-by-step explanation:

\cos( \beta  )  +  \sin( \beta )  \tan( \beta )  =  \sec( \beta )

\cos( \beta )  +  \sin( \beta )  \frac{ \sin( \beta ) }{ \cos( \beta ) }

\cos( \beta )  +  \frac{ \sin {}^{2} ( \beta ) }{ \cos( \beta ) }

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3 0
3 years ago
Read 2 more answers
The large piston in a hydraulic lift has an area of 2m^2. What force must be applied to the small piston with an area of .2m^2 i
Helen [10]

Answer:

147,000N

Step-by-step explanation:

A1= 2m^2

A2= 0.2m^2

F2= 14,700N

Required

F1, the applied force

Applying the formula

F1/A1= F2/A2

substute

F1/2=14700/0.2

2*14700= F1*0.2

29400= F1*0.2

F1= 29400/0.2

F1=147,000N

Hence, the applied force is 147,000N

3 0
3 years ago
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