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Rainbow [258]
2 years ago
12

A dolphin swims 665,280 feet in 3 hours. Use the formula d = rt, where d represents distance, r represents rate, and t represent

s time, to answer the following questions. Show your work.
Part A: Rearrange the distance formula, d = rt, to solve for rate.

Part B: Find the dolphin's rate in feet per hour.

Part C: Find the dolphin's rate in miles per hour.

Part D: Which unit, feet, or miles, makes more sense to use in this scenario, and why?
Mathematics
1 answer:
liraira [26]2 years ago
6 0

Answer:

Answers Below

Step-by-step explanation:

Part A:

r=d/t

Part B:

r=665280/3

r=221760

Part C:

221760/5280 = 42mph

Part D: Miles per Hour, because it is easier to understand

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A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^
guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj  .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j

B )

Acceleration a is given by :

a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j

Hence , this is the required solution .

5 0
3 years ago
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
aliya0001 [1]

The Lagrangian

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)

has critical points where the first derivatives vanish:

L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}

L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}

L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}

L_\lambda=x^4+y^4+z^4-13=0

We can't have x=y=z=0, since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of \pm\sqrt[4]{13}. For example, if y=z=0, then x^4=13\implies x=\pm\sqrt[4]{13}.

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find \lambda=-\frac1{\sqrt{26}} (taking the negative root because x^2,y^2,z^2 must be non-negative), and we can immediately find the critical points from there. For example, if z=0, then x^4+y^4=13. If both x,y are non-zero, then x^2=y^2=-\frac1{2\lambda}, and

xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}

\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}

and for either choice of x, we can independently choose from y=\pm\sqrt[4]{\frac{13}2}.

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then x^2=y^2=z^2=-\frac1{2\lambda}. We have

xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}

\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}

and similary y,z have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate f at each critical point; you should end up with a maximum value of \sqrt{39} and a minimum value of \sqrt{13} (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

(\sqrt[4]{13},0,0)

(-\sqrt[4]{13},0,0)

(0,\sqrt[4]{13},0)

(0,-\sqrt[4]{13},0)

(0,0,\sqrt[4]{13})

(0,0,-\sqrt[4]{13})

\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

5 0
3 years ago
If g(x)= x^2-2x-6 and h(x)= 2x^2-5x+2 find (h-g) (-2)
Natalka [10]

Answer:

<h2>(h - g)( - 2) = 18</h2>

Step-by-step explanation:

g(x) = x² - 2x - 6

h(x) = 2x² - 5x + 2

To find (h-g) (-2) we must first find h - g(x)

To find h - g(x) subtract g(x) from h(x)

We have

<h3>(h - g)(x) =  {2x}^{2}  - 5x + 2 - ( {x}^{2}  - 2x - 6)  \\  =  {2x}^{2}  - 5x + 2 -  {x}^{2}  + 2x + 6 \\  =  {2x}^{2}  -  {x}^{2}  - 5x + 2x + 2 + 6</h3><h3>(h - g)(x) =  {x}^{2}  - 3x + 8</h3>

To find (h-g) (-2) substitute the value of x that's - 2 into (h - g)(x) that's replace every x in (h - g)(x) by - 2

That's

<h3>(h - g)( - 2) =  ({ - 2})^{2}  - 3( - 2) + 8 \\  = 4 + 6 + 8 \\  = 10 + 8</h3>

We have the final answer as

<h3>(h - g)( - 2) = 18</h3>

Hope this helps you

7 0
3 years ago
Laura has $30 to spend on food for the week. Which inequality represents this? x &lt; 30 x &lt; 30 x &gt; 30 x &gt; 30
Llana [10]
X is less than or equal to $30, she can spend $30, she cant spend more, she can spend less.

The answer should look like X<30 with a line under the less than sign, i hope this helps.
7 0
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At most, how many unique roots will a third-degree polynomial function have?
beks73 [17]

The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.

In fact, if you use complex number, then a polynomial with degree n has exactly n roots.

So, in particular, a third-degree polynomial can have at most 3 roots.

In fact, in general, if the polynomial p(x) has solutions x_1,\ x_2,\ldots x_n, then you can factor it as

p(x) = (x-x_1)(x-x_2)\ldots (x-x_n)

So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as

p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)

But this is a fourth-degree polynomial.

7 0
3 years ago
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