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dangina [55]
4 years ago
9

What are 2 consecutive intigers who sum 547?

Mathematics
1 answer:
Elodia [21]4 years ago
4 0
A + b = 547
 b= a +1
 a + a+1 = 547
2a= 547-1
2a= 546
 a= 273
b= 273+1= 274

a= 273   b= 274
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I think it’s 55 hours, about more than 2 days
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A gymnasium is 88 feet long and 76 feet wide. On a blueprint, the gymnasium is 5.5 inches long. What is the width of the gymnasi
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How do you solve this limit of a function math problem? ​
hram777 [196]

If you know that

e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving e.

For starters, we have

\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}

Let y=\dfrac34(x+1). Then as x\to\infty, we also have y\to\infty, and

2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2

So in terms of y, the limit is equivalent to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}

Now use some of the properties of limits: the above is the same as

\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}

The first limit is trivial; \dfrac1y\to0, so its value is 1. The second limit comes out to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}

To see why this is the case, replace y=-z, so that z\to-\infty as y\to\infty, and

\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e

Then the limit we're talking about has a value of

\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}

# # #

Another way to do this without knowing the definition of e as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)

(where the notation means \exp(x)=e^x, just to get everything on one line).

Recall that

\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)

if f is continuous at x=c. \exp(x) is continuous everywhere, so we have

\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)

For the remaining limit, write

\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}

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\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83

and our original limit comes out to the same value as before, \exp\left(-\frac83\right)=\boxed{e^{-8/3}}.

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3 years ago
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Read 2 more answers
Which equation is y = 2x2 – 8x 9 rewritten in vertex form? y = 2(x – 2)2 9 y = 2(x – 2)2 5 y = 2(x – 2)2 1 y = 2(x – 2)2 17
saveliy_v [14]

Answer:

Step-by-step explanation:

we have

y=2x2-8x+9

Group terms that contain the same variable, and move the constant to the opposite side of the equation

y-9=2x2-8x

y-9=2(x2-4x)

y-9+8=2(x2-4x+4)

y-1=2(x2-4x+4)

y-1=2(x-2)2

y=2(x-2)2+1

the answer is the option C

y=2(x-2)2+1

3 0
2 years ago
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