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DochEvi [55]
3 years ago
13

Please help if you can i really need it

Mathematics
1 answer:
Mila [183]3 years ago
4 0

Answer:

9

Step-by-step explanation:

60 requests are made.

Spencer gets 25% of these requests. Convert this to a decmal (divide by 100) to get 0.25. 25% of 60 is:

(0.25)(60)=15\\

So Spencer gets 15 requests. Now do the same process for Tanvi:

(0.10)(60)=6

So Tanvi gets 6 requests. To find out how many more requests Spencer gets compared to Tanvi:

15-6=9

Therefore, Spencer gets 9 more requests than Tanvi.

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A tree is 44.55 feet tall. What is its height in meters?
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PLEASE HELP!!!!
rjkz [21]
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4 years ago
HELPPP PLEASE ILL GIVE BRAINLIEST
kirza4 [7]

Answer:

D. 13 x 9

Step-by-step explanation:

12 11/16 is closer to 13 than 12.

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3 0
3 years ago
Find the fifth roots of 32(cos 280° + i sin 280°).
ikadub [295]
<span>Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4

</span><span>First 5th root: k = 0
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
  (32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
  (32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
   2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
  2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]
  2*[ cos( 280/5 ) + i*sin( 280/5 ) ]
  2*[ cos( 56 ) + i*sin( 56 ) ]
   -------------------------------------------------------------------
 Second 5th root: k = 1
  r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
  (32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
  (32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
  2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
  2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]
  2*[ cos( 640/5 ) + i*sin( 640/5 ) ]
  2*[ cos( 128 ) + i*sin( 128 ) ]
 -------------------------------------------------------------------
 Third 5th root: k = 2
  r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
  (32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
  (32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
  2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
   2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]
  2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]
  2*[ cos( 200 ) + i*sin( 200 ) ]
  -------------------------------------------------------------------
 Fourth 5th root: k = 3
  r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
  (32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
  (32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
  2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
  2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]
  2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]
  2*[ cos( 272 ) + i*sin( 272 ) ]
   -------------------------------------------------------------------
 Fifth 5th root: k = 4
  r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
  (32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
  (32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
  2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
  2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]
  2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]
  2*[ cos( 344 ) + i*sin( 344 ) ]</span>
3 0
4 years ago
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