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3 years ago

Find the value of w if the expression wx(3y² + 6y – 2) simplifies to the expression 6xy² + 12xy – 4x.

Mathematics

2 answers:

tatyana61 [14]3 years ago

6 0

Answer:

-7

Step-by-step explanation:

nadezda [96]3 years ago

3 0

Answer:

w = 2

Step-by-step explanation:

Distribute the expression and compare like terms with the simplified version.

Given

wx(3y² + 6y - 2) ← distribute parenthesis

= 3wxy² + 6wxy - 2wx

Compare coefficients of like terms with

6xy² + 12xy - 4x

Compare xy² term, then

3w = 6 ( divide both sides by 3 )

w = 2

Compare xy term, then

6w = 12 ( divide both sides by 6 )

w = 2

Compare x term, then

- 2w = - 4 ( divide both sides by - 2 )

w = 2

Hence the required value of w is 2

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2 years ago

jennifer bikes 7 miles south and then turns to bike 13 miles east how far away is she from where she started?

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Answer:

She is about 14.765 miles ( $\sqrt{218}$ miles) from where she started

Step-by-step explanation:

There is a relation between the three sides of the right triangle

The side opposite to the right angle is called hypotenuse and it is the longest side

The other two sides called legs of the right angle

The relation between them is: (hypotenuse)² = (leg1)² + (leg2)²

∵ Jennifer bikes 7 miles south

∵ She turns to bike 13 miles east

∵ South and East are perpendicular

→ That means the distance from her start point to end point represents

a hypotenuse of a right triangle, whose legs are 7 and 13

∴ (hypotenuse)² = (leg1)² + (leg2)², where

hypotenuse is the distance between her start and end points

leg1 is her distance in south direction

leg2 is her distance in east direction

∵ Leg1 = 7 miles

∵ leg 2 = 13 miles

∴ (hypotenuse)² = (7)² + (13)²

∴ (hypotenuse)² = 49 + 169

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3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

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Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

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