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Dima020 [189]
3 years ago
7

An alligator can eat 18 pounds of meat in 2 hours. If it continues eating at this same rate, how much can it eat in 7 hours?

Mathematics
1 answer:
Mariana [72]3 years ago
5 0

Answer:

63 pounds

Step-by-step explanation:

Given data

Quantity of meat= 18 pounds

TIme taken= 2 hours

Rate= Quantity/Time taken

Rate= 18/2

Rate= 9 pounds per hour

Now time= 7 hours

Rate= 9 pounds per hour

Quantity= Rate*Time

Quantity= 9*7

Quantity= 63 pounds

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Solve for x . Leave your answer in the simplest form
Alisiya [41]

Hello!

Here we are given a composite shape, meaning that we can divide this shape up into ways that we are given formulas to solve for legs and such.

I can see here that this can be divided into a triangle and a rectangle using a horizontal line.

The leg lengths of this triangle would be 9, because it is congruent to the bottom side of the rectangle, and 7, because it is the left side minus the right side.

This would also make a right triangle, meaning we could solve for the hypotenuse utilizing the Pythagorean Theorem.

a^2+b^2=c^2

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3 years ago
NO LINKS OR FILES!
Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

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Answer:

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