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3 years ago

It's Snowing in Syracuse

Mathematics

1 answer:

larisa86 [58]3 years ago

8 0

Answer:

13 inches ; 2 inches per hour ; 48 inches

Step-by-step explanation:

Given that :

Inches of snow after 2 hours = 4 inches

Since, snow is falling steadily ;

We can infer that:

Inches of snow per hour : (4 in / 2 hrs) = 2 in / hr

Hence, rate of snow fall = 2in /hr

Inches of snow after 6.5 hours

Rate per hour * Numbee of hours

2 * 6.5

= 13 inches

Amount of snow after x hours :

Rate * x

2 * x

= 2x

Inches of snow after 24 hours

2 * 24 hrs = 48 inches

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2 years ago

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In the morning Mary walked 2 1/4 miles around the park. In the afternoon she walked another 5 7/8. how many miles in total did s

anygoal [31]

Answer:

8 1/8 total miles walked.

Step-by-step explanation:

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3 years ago

the density of a substance is 2.5 cm calculate the volume of the substance if its mass is 25.0 grams

sesenic [268]

Answer:

10 cm³

Step-by-step explanation:

The equation for density is: d = m/v, where d is the density, m is the mass, and v is the volume.

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2 years ago

Three times a number h is at least -12 .

pashok25 [27]

Answer:

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3 years ago

If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

   dy
Find —— for an implicit function:
    dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

   dy
Now, isolate —— in the equation above:
   dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

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2 years ago