Find the equation of OP.

Pls help asap or else my parents will kill me eeeeeasy and explian for brainleest .

gtnhenbr [62]

Answer:

Step-by-step explanation:

I'm assuming that your "70=50x what is xx" actually means "find x if 70-50x = 0" If that's not what you meant, please retype your question.

Dividing both sides by -50 results in -70/50 + x = 0, and adding 70/50 to both sides results in x = 70/50, or 7/5.

Only if my assumptions are correct: x = 7/5.

3 0

2 years ago

50 POINTS

MatroZZZ [7]

Answer:

see below

Step-by-step explanation:

(x) = 7 - 2x

Let t(x) =0

0 = 7-2x

Subtract 7 from each side

-7 = -2x

Divide by -2

-7/-2 =-2x/-2

7/2 =x

h(x) = 4x + 2

Replace x with x+3

h(x+3) = 4(x+3) + 2

Distribute

= 4x+12 +2

=4x+14

3 0

3 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago

Plss help!!! (15pts)

liubo4ka [24]

Answer:

w=54

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

ATH CHALLENGE

romanna [79]

Answer:

78

Step-by-step explanation:

for the right side we can just do 14*2 since there are two people per seat and 14 rows

14 *2 is 28

for the left side we just do 14*3 since there are three people per seat and 14 rows

14*3 is 42

Then we have to add the 8 students that can fit in the very back seat

42 + 28 + 8

28 + 8 is 36

42 + 36 = 78

3 0

2 years ago