Whakaoti i te (t)=40t-5t^2

$$( t ) = 40 t - 5 t ^ { 2 }$$

Whakaoti mō t

$$t = \frac{39}{5} = 7\frac{4}{5} = 7.8$$

$$t=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 40t mai i ngā taha e rua.

$$t-40t=-5t^{2}$$

Pahekotia te t me -40t, ka -39t.

$$-39t=-5t^{2}$$

Me tāpiri te 5t^{2} ki ngā taha e rua.

$$-39t+5t^{2}=0$$

Tauwehea te t.

$$t\left(-39+5t\right)=0$$

Hei kimi otinga whārite, me whakaoti te t=0 me te -39+5t=0.

$$t=0 t=\frac{39}{5}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 40t mai i ngā taha e rua.

$$t-40t=-5t^{2}$$

Pahekotia te t me -40t, ka -39t.

$$-39t=-5t^{2}$$

Me tāpiri te 5t^{2} ki ngā taha e rua.

$$-39t+5t^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$5t^{2}-39t=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -39 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}}}{2\times 5}$$

Tuhia te pūtakerua o te \left(-39\right)^{2}.

$$t=\frac{-\left(-39\right)±39}{2\times 5}$$

Ko te tauaro o -39 ko 39.

$$t=\frac{39±39}{2\times 5}$$

Whakareatia 2 ki te 5.

$$t=\frac{39±39}{10}$$

Nā, me whakaoti te whārite t=\frac{39±39}{10} ina he tāpiri te ±. Tāpiri 39 ki te 39.

$$t=\frac{78}{10}$$

Whakahekea te hautanga \frac{78}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$t=\frac{39}{5}$$

Nā, me whakaoti te whārite t=\frac{39±39}{10} ina he tango te ±. Tango 39 mai i 39.

$$t=\frac{0}{10}$$

Whakawehe 0 ki te 10.

$$t=0$$

Kua oti te whārite te whakatau.

$$t=\frac{39}{5} t=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 40t mai i ngā taha e rua.

$$t-40t=-5t^{2}$$

Pahekotia te t me -40t, ka -39t.

$$-39t=-5t^{2}$$

Me tāpiri te 5t^{2} ki ngā taha e rua.

$$-39t+5t^{2}=0$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$5t^{2}-39t=0$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5t^{2}-39t}{5}=\frac{0}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$t^{2}-\frac{39}{5}t=\frac{0}{5}$$

Whakawehe 0 ki te 5.

$$t^{2}-\frac{39}{5}t=0$$

Whakawehea te -\frac{39}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{39}{10}. Nā, tāpiria te pūrua o te -\frac{39}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}-\frac{39}{5}t+\left(-\frac{39}{10}\right)^{2}=\left(-\frac{39}{10}\right)^{2}$$

Pūruatia -\frac{39}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}-\frac{39}{5}t+\frac{1521}{100}=\frac{1521}{100}$$

Tauwehea t^{2}-\frac{39}{5}t+\frac{1521}{100}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{39}{10}\right)^{2}=\frac{1521}{100}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(t-\frac{39}{10}\right)^{2}}=\sqrt{\frac{1521}{100}}$$

Whakarūnātia.

$$t-\frac{39}{10}=\frac{39}{10} t-\frac{39}{10}=-\frac{39}{10}$$

Me tāpiri \frac{39}{10} ki ngā taha e rua o te whārite.

$$t=\frac{39}{5} t=0$$