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4 years ago

Evaluate the following expression using the values given:

Mathematics

1 answer:

Free_Kalibri [48]4 years ago

7 0

x^3 − 2y^2 − 3x^3 + z^4

Plug in 3 for x, 5 for y, and -3 for z (given)

(3)^3 -2(5^2) - 3(3^3) + (-3)^4

Simplify

(3)^3 = 3 x 3 x 3 = 27

-2(5^2) = -2 x 25 = -50

-3(3^3) = -3 x 27 = -81

(-3)^4 = -3 x -3 x -3 x -3 = 9 x 9 = 81

Combine all the terms

27 + (-50) - 81 + 81

27 - 50 - 81 + 81

-23 + 0

-23

-23, or (C) is your answer

hope this helps

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What is the solution to 5/6x-1/3>1 1/3

CaHeK987 [17]

Answer:

Step-by-step explanation:

$\frac{5}{6}x-1>1\frac{1}{3}\\\\\frac{5}{6}x-1>\frac{4}{3}\\\\\frac{5}{6}x>\frac{4}{3}+1\\\\\frac{5}{6}x>\frac{4}{3}+\frac{3}{3}\\\\\frac{5}{6}x>\frac{7}{3}\\\\x>\frac{7}{3}*\frac{6}{5}\\\\x>\frac{7*2}{5}\\x>\frac{14}{5}\\\\x>2\frac{4}{5}$

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3 years ago

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bija089 [108]

X+93=4x *vertical angles are congruent*
93=3x
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6 0

3 years ago

You have learned several methods for solving a system of equations. First, rank the methods in order of preference, noting which

kifflom [539]

Methods for solving a system of equations:
- elimination
- substitution
- augmented matrix

Ranks of method ( number one being the most preferred method)
1) elimination
2) substitution
3) augmented matrix

Elimination is my preferred method because it is simple and easy to work out and hard to get wrong.

Augmented matrix is my least preferred method because it is easy to get wrong and it requires complex steps to solve a simple equation.

Hope this helps ;)

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3 years ago

Type the correct answer in the box. Spell all words correctly.

MA_775_DIABLO [31]

Answer:

product

Step-by-step explanation:

the solution or answer to a multiplication problem is a product.

hope this helps have a good day and stay safe :)

7 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

4 years ago