proszę o pomoc z drugim zadaniem dam naj i 30 pkt


Proszę O Pomoc Z Drugim Zadaniem Dam Naj I 30 Pkt class=

Odpowiedź :

Odpowiedź:

A.

[tex] \frac{3}{8} [/tex]

B.

[tex] - 2 \frac{2}{5} [/tex]

C.

[tex] - 200.08[/tex]

D.

[tex]-30[/tex]

Szczegółowe wyjaśnienie:

A.

[tex] - {( - \frac{1}{2} )}^{3} + {( \frac{1}{2}) }^{2} = - ( - \frac{1}{8} ) + \frac{1}{4} = \frac{ 1}{8} + \frac{2}{8} = \frac{3}{8} [/tex]

B.

[tex] {( \frac{4}{5} )}^{2} - \frac{ {4}^{2} }{5} + \frac{4}{ {5}^{2} } = \frac{16}{25} - \frac{16}{5} + \frac{4}{25} = \frac{20}{25} - \frac{16}{5} = \frac{4}{5} - \frac{16}{5} = - \frac{12}{5} = - 2 \frac{2}{5} [/tex]

C.

[tex]10 \times {( - 0.2)}^{3} - 0.2 \times {10}^{3} = 10 \times - 0.08 - 0.2 \times 1000 = - 0.8 - 200 = - 200.08[/tex]

D.

[tex] {6}^{2} \times \frac{1}{6} - { ( - \frac{1}{6} )}^{0} \times {( - 6)}^{2} = 6 - 1 \times 36 = - 30[/tex]

W razie pytań pisz śmiało,

Venty

Odpowiedź:

[tex] a) \: - ( - \frac{3}{2} {)}^{3} + ( \frac{1}{2} {)}^{2} = - ( - \frac{27}{8} ) + \frac{1}{4} = \frac{27}{8} + \frac{2}{8} = \frac{29}{8} = 3 \frac{5}{8} [/tex]

[tex]b) \: \: ( \frac{4}{5} {)}^{2} - \frac{ {4}^{2} }{5} + \frac{4}{ {5}^{2} } = \frac{16}{25} - \frac{16}{5} + \frac{4}{25} = \frac{16}{25} - \frac{90}{25} + \frac{4}{25} = - \frac{78}{25} = - 3 \frac{3}{25} [/tex]

[tex]c) \: \: 10 \times ( - 0.2 {)}^{3} - 0.2 \times {10}^{3} = 10 \times ( - 0.008) - 0.2 \times 1000 = - 0.08 - 200 = - 200.08[/tex]