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2 years ago

Suppose that a duck swam 22.5 kilometers in 5 hours at a constant speed how far did the duck travel

Mathematics

2 answers:

SVEN [57.7K]2 years ago

6 0

22.5 km divided by 5 hours equals a constant speed of 4.5 km per hour.

jekas [21]2 years ago

3 0

Answer:

4.5 = x

Step-by-step explanation:

Given:

22.5 = 5x

(we need to find out what x is)

22.5 / 5 = 4.5

meaning that the duck traveled at 4.5 kilometers per hour.

Check:
4.5 x 5 = 22.5

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What is an efficient way to compute 47^2–37^2 without using raw calculations?

e-lub [12.9K]

Answer:

(47+37)(47-37)

Step-by-step explanation:

47^2-37^2 is a difference of two squares. a^2-b^2 = (a+b)(a-b) so this expression can be factored to (47+37)(47-37) = (84)(10) = 840

6 0

2 years ago

Matt and Kelly both purchased the same type mulch to be used in a garden. They both spent exactly $60 each on mulch. • Matt purc

hichkok12 [17]

Answer:

B.) Kelly got the better deal.

Step-by-step explanation:

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3 years ago

(Look at the picture) help me

umka21 [38]

Answer:

i had this in rsm a long time ago, i think the answer(if it's an equation) should be:

6x + x = 7x

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Step-by-step explanation:

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2 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

7 0

3 years ago

Find the volume of the cylinder to the nearest tenth use 3.14

Maurinko [17]

the answer is 3.14 r^2 h

the h is the height

3 0

2 years ago

Read 2 more answers