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2 years ago

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At time t is greater than or equal to zero, a cube has volume V(t) and edges of length x(t). If the volume of the cube decreases

at a rate proportional to its surface area, which of the following differential equations could describe the rate at which the volume of the cube decreases?
A) dV/dt=-1.2x^2
B) dV/dt=-1.2x^3
C) dV/dt=-1.2x^2(t)
D) dV/dt=-1.2t^2
E) fav/dt=-1.2V^2

2 answers:

algol [13]2 years ago

8 0

Answer:

A) dV/dt=-1.2x^2

Step-by-step explanation:

The rate of change of volume is given by dV/dt. Surface area is proportional to x^2. Since the volume is decreasing, the constant of proportionality between surface area and rate of volume change will be negative. Hence a possible equation might be ...

dV/dt = -1.2x^2

VARVARA [1.3K]2 years ago

6 0

Answer:

C

Step-by-step explanation:

V(t) = [x(t)]³

A(t) = 6[x(t)]²

dV/dt = k × 6[x(t)]²

Where k < 0

From the options,

taking k = -0.2

dV/dt = -1.2[x(t)]²

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Answer: (-2, -5), (-1, -8), (0, -9), (1, -8), (2, -5)

Step-by-step explanation:

f(-2)=(-2)^2-9

f(-2)=4-9

f(-2)=-5 ==> (-2, -5)

f(-1)=(-1)^2-9

f(-1)=1-9

f(-1)=-8 ==> (-1, -8)

f(0)=0^2-9

f(0)=0-9

f(0)=-9 ==> (0, -9)

f(1)=(1)^2-9

f(1)=1-9

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2 years ago

I need the answers to this please and thank you:)

Liono4ka [1.6K]

Q # 1

Explanation

Given the parabola

$f\left(x\right)=\left(x-3\right)^2-1$

Openness

It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2-6x+8$

$a=1,\:b=-6,\:c=8$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

Finding $y_v$

$y_v=3^2-6\cdot \:3+8$

$y_v=-1$

So vertex is:

$\left(3,\:-1\right)$

Horizontal Translation

$y=\left(x-3\right)^2$ moves the graph RIGHT 3 units.

Vertical Translation

$f\left(x\right)=\left(x-3\right)^2-1$ moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Q # 2

Explanation:

$f\left(x\right)=-\left(x+1\right)^2-2$

Openness

It OPENS DOWN, as 'a=-1' is negative.

Vertex

$\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=-x^2-2x-3$

$a=-1,\:b=-2,\:c=-3$

$x_v=-\frac{\left(-2\right)}{2\left(-1\right)}$

$x_v=-1$

$\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-2$

So vertex is:

$\left(-1,\:-2\right)$

Horizontal Translation

$y=\left(x+1\right)^2$ moves the graph LEFT 1 unit.

Vertical Translation

$f\left(x\right)=\left(x+1\right)^2-2$ moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As $a = -1$ < 0, so it is either stretched or compressed.

Q # 3

Explanation:

$f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6$

It OPENS UP, as 'a=1/3' is positive.

Vertex

$\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}$

$a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}$

$x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}$

$x_v=4$

Finding $y_v$

$y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}$

$y_v=6$

So vertex is:

$\left(4,\:6\right)$

Horizontal Translation

$f\left(x\right)=\left(x-4\right)^2$ moves the graph RIGHT 4 units.

Vertical Translation

$f\left(x\right)=}\left(x-4\right)^2+6$ moves the graph UP 6 unit.

Stretch or Compress Vertically

As $a=\frac{1}{3}$ , so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

Q # 4

Explanation:

Given the function

$f\left(x\right)=-\left(x+3\right)^2$

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form $y=a\left(x-m\right)\left(x-n\right)$

is the average of the zeros $x_v=\frac{m+n}{2}$

$y=-\left(x+3\right)^2$

$a=-1,\:m=-3,\:n=-3$

$x_v=\frac{m+n}{2}$

$x_v=\frac{\left(-3\right)+\left(-3\right)}{2}$

$x_v=-3$

Finding $y_v$

$y_v=-\left(-3+3\right)^2$

$y_v=0$

So vertex is:

$\left(-3,\:0\right)$

Horizontal Translation

$y=\left(x+3\right)^2$ moves the graph LEFT 3 units.

Vertical Translation

$y=\left(x+3\right)^2$ does not move the graph vertically.

Stretch or Compress Vertically

As $a=-1$ , so it the graph is either vertically stretched or compressed.

Q # 5

Explanation:

$f\left(x\right)=\left(x+5\right)^2-3$

Openness

It OPENS UP, as 'a=1' is positive.

Vertex

$\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2+10x+22$

$a=1,\:b=10,\:c=22$

$x_v=-\frac{10}{2\cdot \:1}$

$x_v=-5$

Finding $y_v$

$y_v=\left(-5\right)^2+10\left(-5\right)+22$

So vertex is:

$\left(-5,\:-3\right)$

Horizontal Translation

$f\left(x\right)=\left(x+5\right)^2$ moves the graph LEFT 5 units.

Vertical Translation

$f\left(x\right)=\left(x+5\right)^2-3$ moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

Q # 6

THE DETAILS OF COMPLETE SOLUTION OF QUESTION 6 IS ATTACHED IN THE DIAGRAM AS THE 5000 CHARACTERS WERE ALREADY FILLED. SO, I solved via the attached figure.

SO, PLEASE CHECK THE LAST FIGURE TO FIND THE COMPLETE SOLUTION OF THE Q#6.

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3 years ago

Polygon A B C D E F has 6 sides. Exterior angle A is 62 degrees, exterior angle B is 44 degrees, exterior angle C is 70 degrees,

nevsk [136]

The measure of angle AFE would be 99 degrees.

Step-by-step explanation:

Given that,

Exterior angle A = 62 degrees

Exterior angle B = 44 degrees

Exterior angle C = 70 degrees

Exterior angle D = 60 degrees

Exterior angle E = 43 degrees

Angle AFE = ?

However, for that, we have to find the exterior angle of F.

We know that the sum of exterior angles of polygon is 360 degrees.

Exterior angle( A + B + C + D + E + F) = 360 degrees

Exterior angle F = 360 - Exterior angle( A + B + C + D + E )

Exterior angle F = 360 - (62 +44 + 70 + 60 + 43 )

Exterior angle F = 360 - 279

Exterior angle F = 81

The, sum of interior and exterior angle is 180 degrees. Therefore, the angle AFE would be 180 - 81 = 99 degrees.

Keywords: polygon, exterior angle

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2 years ago

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Can two numbers have a GCF that is bigger than their LCM? Explain how you know?

pshichka [43]

The answer to this question would be: No, they can't

Greatest common factor biggest possible value is the number value itself. There is no factor that bigger than the number.

But the least common multiple smallest possible value is the number. There is no multiple that smaller than the number.

So it is not possible to have GCF bigger than LCM.

3 0

3 years ago