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sp2606 [1]
1 year ago
15

I need help please Asap

Mathematics
1 answer:
s2008m [1.1K]1 year ago
4 0

Answer:

Step-by-step explanation:

1.

Equation one:

x = -5, x = -1 (Both are real)

Equation two:

No real solutions

Equation three:

x = -3 (Real)

Equation four:

No real solutions

2.

The easiest way to figure out if an equation has real solutions is to factor it. If it is factorable, then it has real solutions. If it isn't, then it doesn't have real solutions.

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xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}

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Story : A Dog’s Tale by Mark Twain

4. Read this sentence from the selection.

“Whenever she heard a large word she said it over to herself many times, and so was able to keep it until there was a dogmatic gathering in the neighborhood.”

In this sentence the word dogmatic is used as —

• an insightful simile
• a descriptive metaphor
• a humorous play on words
• an illuminating allusion


5. In the last paragraph the narrator says his mother would “fetch out a long word.” The connotations of the word fetch remind the reader that the narrator’s mother —

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