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2 years ago

Use two unit multipliers to convert 20 inches to miles

Mathematics

1 answer:

jarptica [38.1K]2 years ago

6 0

Answer:

Step-by-step explanation:

I answered another one of your questions so I will not explain as much. ther are 12 inches in a foot and then 5280 feet in a mile so that's 1 ft/ 12 in and 1 mi/ 5280 ft. So the math is 20 in * 1 ft/ 12 in * 1 mi/ 5280 ft = .000315 mi.

In reference to the last one, do keep in mind "square" inches and whatnot mean you square things. so 1 ft/ 12 in becomes 1 ft^2/ 144 in^2. Just as a small heads up to what could be tricky.

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Preston and Joel are both solving the equation 2x=14. Preston is not sure what to do because he does not know a power of 2 that

Murrr4er [49]

Answer:

yes

Step-by-step explanation:

You can always separate an equation into two parts and see where those graphs intersect.

Joel's method works well.

_____

Additional comments

Preston should know that the invention of logarithms makes it easy to solve equations like this. x = log₂(14) = log(14)/log(2) ≈ 3.8073549.

As for Joel's method, I prefer to subtract the right side to get the equation ...

2^x -14 = 0

Then graphing y = 2^x -14, I look for the x-intercept. Most graphing calculators make it easy to find x- and y-intercepts. Not all make it easy to find points of intersection between different curves.

7 0

2 years ago

Which graph decreases crosses the y axis at (0, -7) and then remains constant

deff fn [24]

Answer:

make a line going slash down to -7 and then draw a horizontial line parrel to the x axis

Step-by-step explanation:

here is a link to model rebrand.ly/s3gep1k

8 0

2 years ago

Find dy/dy in terms of x and y

dmitriy555 [2]

$\frac{dy}{dy}=1$ , so I assume you mean "find $\frac{dy}{dx}$ ".

We can rewrite this as an implicit equation to avoid using too much of the chain rule, namely

$y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} \implies (x^2+1) y^3 = e^x (x+1)$

Differentiate both sides with respect to $x$ using the product and chain rules.

$2x y^3 + 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x(x+1) + e^x$

$\implies 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x (x+2) - 2x y^3$

$\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x y^3}{3(x^2+1) y^2}$

Now substitute the original expression for $y$ .

$\dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^3}{3(x^2+1) \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^2}$

$\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - \frac{2e^x(x^2+x)}{x^2+1}}{3(x^2+1) \left(\frac{e^x(x+1)}{x^2+1}\right)^{2/3}}$

$\implies \dfrac{dy}{dx} = e^x \dfrac{x^3-x+2}{3(x^2+1)^2 \frac{e^{2x/3}(x+1)^{2/3}}{(x^2+1)^{2/3}}}}$

$\implies \dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}}$

Now, since

$y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}}$

we can write

$\dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}} \times \dfrac{x^3-x+2}{3(x^2+1)^{3/3} (x+1)^{3/3}}$