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pav-90 [236]
3 years ago
10

Answer this simple quick question!! Pls :) m

Mathematics
2 answers:
notka56 [123]3 years ago
7 0

Answer:

Answer is B!

Step-by-step explanation:

58 * 20 = 1160

0+160= 160

160+1000=1160

Olenka [21]3 years ago
6 0

Answer:

B

Step-by-step explanation:

58 * 20 is 1160 and 0 + 160 + 1000 = 1160

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How many solutions does the equation x_1 +x_2+x_3+x_4+x_5=21 have where x_1, x_2, x_3, x_4, and x_5 are nonnegative integers and
omeli [17]

Step-by-step explanation:

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 21\\     (given)

Let us consider :

x_{1} = t_{1} + 1

x_{2} = t_{2}

x_{3} = t_{3}

x_{4}  = t_{4}

x_{5} = t_{5}

Now, by substituting the above considerations in the above equation, we get:

t_{1} + 1 + t_{2} + t_{3} + t_{4} + t_{5} = 21\\

t_{1}  + t_{2} + t_{3} + t_{4} + t_{5} = 20\\

where,

t_{i} \geq 1

then it follows

n = 20

r = 4

then no. of solutions for the eqn = _{r}^{n + r}\textrm{C}

                                                      = _{4}^{24}\textrm{C}

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Answer :

no. of solutions for the eqn 10626

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