2 years ago

Rezolvati, in multimea numerelor intregi, inecuatiile Exercitiul 8 va rog!

Mathematics

1 answer:

geniusboy [140]2 years ago

7 0

Answer: hi

Step-by-step explanation:

what the hell does that supposed to mean

i cant under stand S H I T we speak English not Japanese

people are so dumb and stupid these days

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find three numbers whose sum is 32 if the second is 1 more than the first and the first is 4 less than the third

Alex777 [14]

X+y+z=32
x=z-4
y=x+1
z=x+4
x+x+1+x+4=32
3x+5=32
3x=27
x=9
y=10
z=13

Hope this helps :)

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2 years ago

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Ksivusya [100]

Answer:

Step-by-step explanation:

$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$