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3 years ago

10

Hi, here is a hotline for all of my lgbq+ friends. Hope we have fun here!! !️‍❤️

2 answers:

AVprozaik [17]3 years ago

8 0

Answer:

YOUR AMAZING LSHFHSDF TYSM FOR PUTTING THIS UP SfDJFH

Step-by-step explanation:

riadik2000 [5.3K]3 years ago

7 0

Answer:

This is a really nice thing to do.

Step-by-step explanation:

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Line segment SU is dilated to create S'U' using the dilation rule DQ,2.5. What is the distance, x, between points U' and U?

hodyreva [135]

Answer:

The answer is 6 units! Have a nice day! :D

Step-by-step explanation:

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3 years ago

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What is standard form

nataly862011 [7]

Example 250

word form:
two hundred and fifty

expanded form:
2+5+0

standard form:
250

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Which expression is equivalent to -16 - 7?

lakkis [162]

Step-by-step explanation:

B. -16 + (-7) ...this is the same as -16-7

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3 years ago

#2 Find the slope given the points (6,-12) and (15,-3)*

castortr0y [4]

Greetings :)

To find slope of two points we will need to use this equation: $\frac{y^2-y^1}{x^2-x^1}$

$x^1$ $x^2$ $y^1$ $y^2$

( 6 , -12 ) ( 15 , -3 )

Now let's replace the equation with the numbers. (it will be a fraction).

$\frac{-3-(-12)}{15-12}$ = $\frac{9}{9}$

The equation equals 9 over 9, which also equals 1.

The slope of the line is 1.

4 0

3 years ago

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