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Vaselesa [24]
2 years ago
9

you are allowed to multiply as many 2's and/or as many 5's as you want. What can be the last didget of your result

Mathematics
1 answer:
Levart [38]2 years ago
5 0

Based on the fact that you are allowed to multiply as many 2's and/ or as many 5's as you want, the last digit of your result can be either a 2, 5 or 0.

<h3>What is the last digit of your result?</h3>

If you multiplied a 5 number with another 5 number, you will either get a 0 as the last digit or a 5 as the last digit:

5 x 5 = 25

5 x 10 = 50

5 x 15 = 75

5 x 25 = 125

If you multiply a 5 with a 2, you result would always end in 0:

5 x 2 = 10

5 x 4 = 20

5 x 6 = 30

If you multiply a 2 with another 2, the result would be a 2:

2 x 2 = 4

2 x 4 = 8

2 x 6 = 12

Find out more on operations involving 5s at brainly.com/question/11977981.

#SPJ1

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Liono4ka [1.6K]

Answer:

29Kg

Step-by-step explanation:

P1=2Kg

P2=7Kg

P3=31Kg

P3-P1=29Kg

To find P1, P2 and P3 I started assigning the first prime number, 2, to P1 and tried to assign prime numbers to P2 and P3 so that the sum was 40, increasing them at each step.

I was lucky and I got the result after few steps :-)

3 0
3 years ago
The product of a number x increased by 10 and 4 is 22.
drek231 [11]

x+10+4=22

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x=22-14

x=8

3 0
3 years ago
Plane hkp and plane rkp are two distinct planes. name the intersection of plane hkp and plane rkp.
Zielflug [23.3K]
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3 0
3 years ago
Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3
Serhud [2]

Answer:

e^{3t}(2t+5t^{2})

Step-by-step explanation:

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]

Separate the fraction in a sum:

e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])

The formula for this is:

L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}

Modify the expression to match the formula.

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])

Solve

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )

6 0
3 years ago
Complete the equation of the line whose yyy-intercept is (0,5)(0,5)left parenthesis, 0, comma, 5, right parenthesis and slope is
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3 years ago
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