Can need help again please anyone

Mathematics

1 answer:

Tom [10]3 years ago

7 0

Answer:

Put one point on (0, -4)

Put the other on (3, -2)

Step-by-step explanation:

That's how graphing works

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Simplify (8m2+m+7m2)-(3m-4m3+2m2)

kondaur [170]

$(8m^2+m+7m^2)-(3m-4m^3+2m^2)\\\\=8m^2+m+7m^2-3m+4m^3-2m^2\qquad|\text{combine like terms}\\\\=(4m^3)+(8m^2+7m^2-2m^2)+(m-3m)\\\\=4m^3+13m^2-2m$

4 0

3 years ago

Which expression is equivalent to -3(4c-2) -2c

sergeinik [125]

Answer:

-3(4c-2) -2c = -14c+6

Step-by-step explanation:

We need to find the equivalent expression for -3(4c-2) -2c.

First we open the brackets as follows :

-3(4c-2) -2c

= -12c+6-2c

=-12c-2c+6

=-14c+6

So, the equivalent expression is -14c+6.

7 0

2 years ago

There are currently 17 frogs in a (large) pond. The frog population grows exponentially, tripling every 6 days. How long will it

Tema [17]

Answer:

t=11,11 days

Step-by-step explanation:

F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating $\int\limits^ {} \, dF/F =K\int\limits^ {} \, dt$ ⇒ LnF=Kt+c, $F=ce^{Kt}$ ; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), $F=ce^{K0} = F=c=17; F=17e^{6K} =51$ ⇒ $e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183$ , so $F=17e^{0,183t}$ , now if F=130, t=? we have: