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AlladinOne [14]
3 years ago
13

How do you do number 20?

Mathematics
1 answer:
LenKa [72]3 years ago
4 0

The answer isnin the pictures

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Hi people of earth... I have a question here...
Mekhanik [1.2K]

Hello person who is also from earth :)

6 - -4 = 10


Have a good day!!

8 0
3 years ago
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What is the maximum number of relative extrema a polynomial function can have?
creativ13 [48]
Relative extrema occur where the derivative is zero (at least for your polynomial function). So taking the derivative we get

<span>20<span>x3</span>−3<span>x2</span>+6=0

</span><span> This is a 3rd degree equation, now if we are working with complex numbers this equation is guaranteed to have 3 solutions by the fundamental theorem of algebra. But the number of real roots are 1 which can be found out by using Descartes' rule of signs. So the maximum number of relative extrema are 1.</span>
4 0
3 years ago
Pls help answer this soon!!!!!
vladimir1956 [14]
13a) y = 0.45x

13b) 52 * 0.45 = 23.4

13c) I will let u figure this one out
7 0
3 years ago
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Make r the subject of the formula
zloy xaker [14]

Answer:

r=\frac{-a+p}{a+p}

Step-by-step explanation:

-pr+p=ar+a

-ar-pr+p=a

-ar-pr=a-p

r(-a-p)=a-p

r=\frac{-a+p}{a+p}

3 0
3 years ago
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Imagine the United States only had two postage stamps, a 3 cent stamp and a 7 cent stamp. If you can put any number of stamps on
cricket20 [7]

Answer:

  • not possible:  1, 2, 4, 5, 8, 11
  • possible: all others

Step-by-step explanation:

Obviously, all multiples of 3 and 7 can be paid.

1 cent can be added by adding a 7-cent stamp and taking away two 3-cent stamps. 2 cents can be added by adding three 3-cent stamps and taking away one 7-cent stamp. Hence any value more than 7 +2(3) = 13 cents can always be accommodated.

Amounts that are impossible:

  1, 2, 4, 5, 8, 11 cents

All other positive integer amounts are possible.

7 0
3 years ago
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