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NISA [10]
3 years ago
8

A lake near the Arctic Circle is covered by a 222-meter-thick sheet of ice during the cold winter months. When spring arrives, t

he warm air gradually melts the ice, causing its thickness to decrease at a constant rate. After 333 weeks, the sheet is only 1.251.251, point, 25 meters thick. Let yyy represent the ice sheet's thickness (in meters) after xxx weeks. Complete the equation for the relationship between the thickness and number of weeks.
Mathematics
1 answer:
ioda3 years ago
4 0

A lake near the Arctic Circle is covered by a 2-meter-thick sheet of ice during the cold winter months. When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a constant rate. After 3 weeks, the sheet is only 1.25 meters thick. Let y represent the ice sheet's thickness ( in meters) after x weeks. Complete the equation for the relationship between the thickness and number of weeks.

Answer:

y = 2 - 0.25x

Step-by-step explanation:

From the question, the initial thickness of the ice sheet = 2 meters,

After 3 weeks, the thickness of ice sheet reduced to 1.25 meters

Hence, the difference in the thickness in 3 weeks is calculated as:

2m - 1.25m = 0.75m

The amount of changes that occurred in 3 weeks is given as:

= 0.75/3 = 0.25 meters,

We are told that, the ice is melting with the constant rate.

Therefore, the equation for the relationship between the thickness and number of weeks is given as:

y = 2 - 0.25x

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Which of the following properties would NOT be used to justify any of the steps in solving the equation below?
kykrilka [37]

Given:

The equation is

\dfrac{7}{2}n=3n+4

To find:

The properties which is not used from the given options.

Solution:

We have,

\dfrac{7}{2}n=3n+4

Using Multiplication Property of Equality, multiply both sides by 2.

2(\dfrac{7}{2}n)=2(3n+4)

7n=6n+8                    (Distributive Property)

Using Subtraction Property of Equality, subtract 6n from both sides.

7n-6n=6n+8-6n

n=8

From the given options only one property is not used, i.e., Zero Product Property.

Therefore, the correct option is D.

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3 years ago
Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1
mafiozo [28]

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $ b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value x_{0}  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]

Applying chain rule,

y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

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Identify the slope and y- intercept for the equation y = 3/4 x -2 <br><br> Please help
baherus [9]

Answer:

Slope = 3/4

Y-intercept = -2

Step-by-step explanation:

The equation is already in slope intercept form, y=mx+b

m = slope

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So your slope is 3/4, and your y-intercept is -2!

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