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3 years ago

Write the equation of a line that is perpendicular to the given line and that passes through the given point. y=2/3x+9;(-6,5)

Mathematics

2 answers:

STatiana [176]3 years ago

7 0

The perpendicular equation is y = -3/2x - 4.

You can find this by first realizing that perpendicular lines have opposite and reciprocal slopes. So since it starts at 2/3 we flip it and make it a negative and the new slope is -3/2. Now we can use that and the point to get the y intercept using slope intercept form.

y = mx + b

5 = (-3/2)(-6) + b

5 = 9 + b

-4 = b

And now we can use our new slope and new intercept to model the equation.

y = -3/2x - 4

WARRIOR [948]3 years ago

4 0

Hay dear friend thank you for asking the question

according to the question we know that

y = mx +b

And given that y = 5 , m = -3/2, x = -6

so

5 = -3/2×(-6) +b

5 = 9 + b

-4 = b

now we can use our new slope and new intercept model equation then,

y = (-3/2)x-4

I hope it's help you

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Eva8 [605]

Answer:

b. 1.5

c. 1.5

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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:

$C_1=-1$

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

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

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:

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

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

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3 years ago

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marshall27 [118]

Answer:

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Two pair of vertical angles have been given,

∠ABD and ∠CBE

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One pair of angles ∠ABD and ∠CBE are acute angles which are less than 90°.

Similarly, other pair ∠ABE and ∠DBC are obtuse angles which are more than 90°.

Therefore, out of the given options measure of one pair ∠DBC and ∠ABE should be 135°.

Option (4) will be the answer.

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