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3 years ago

1-1=............... HELP!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Verizon [17]3 years ago

7 0

1-1=0 lol how do u not know this ???

Jobisdone [24]3 years ago

7 0

Answer:

1 + 1 = 1 - 1v = 2 - 2v = 0

Step-by-step explanation:

XD jk lol 1 - 1 is 0 like 1 defeated 1 but the other one got defeated too

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The point P(-6, -6) is reflected over the x-axis. What are the coordinates of the resulting

Fed [463]

Answer:

It would be (6,-6)

Step-by-step explanation:

Hope this helps.

7 0

3 years ago

Which line are parallel

Rufina [12.5K]

Ones that never intersect

4 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

2) Tell whether each statement about the expression 5y + 9 is True or False

olya-2409 [2.1K]

Answer:

True.

Step-by-step explanation:

A variable term is a term with a variable. 5y is your only variable term; therefore, it only has one variable term.

4 0

3 years ago

-6x+5y+2z=-11 -2x+y+4z=-9 4x-5y+5z=-4

AlladinOne [14]

Answer:

$[4, 3, -1]$

Explanation:

When solving systems of equations in three variables, pick two equations to work with first, solve for both variables, plug them into one of the equations you first started on, then in the end, put them altogether:

{-6x + 5y + 2z = -11 ←

{-2x + y + 4z = -9 We can eliminate the y-values obviously

{4x - 5y + 5z = -4 ←

{-6x + 5y + 2z = -11

{4x - 5y + 5z = -4

___________________

-2x + 7z = -15

- 7z -7z

______________

-2x = -15 - 7z

___ ________

-2 -2

x = 7½ + 3½z

-2[7½ + 3½z] + y + 4z = -9

-15 - 7z + y + 4z = -9

-15 - 3z + y = -9

+15 + 15

_______________

-3z + y = 6

+3z + 3z

y = 6 + 3z [Plug in -1 for z to give you the y-value of 3; 3 = y; since you now have both z and y terms, plug in these terms back into all three equations above to get the x-value of 4.]; 4 = x ⤻

-6[7½ + 3½z] + 5[6 + 3z] + 2z = -11

-45 - 21z + 30 + 15z + 2z = -11

-15 - 4z = -11

+15 + 15

_____________

-4z = 4

___ __

-4 -4

z = -1 [Plug this into the equation for y]⤻

I am joyous to assist you anytime.

3 0

3 years ago