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Elena-2011 [213]
3 years ago
6

2.2

Mathematics
1 answer:
kupik [55]3 years ago
5 0

Answer:

2, 5, 11, 23, 47, 95, 191

A, Z, B, Y, C, X, D, W, E, V

Step-by-step explanation:

1. 2, 5, 11, 23, 47, ...95, 191

5 - 2 = 3

12 - 5 = 6

23 - 11 = 12

47 - 23 = 24

Next

(24 × 2) + 47

= 48 + 47

= 95

Next

(48 × 2) + 95

96 + 95

= 191

Next

(96×2) + 191

192 + 191

= 383

The pattern follows (difference ×2) + previous number

That is,

(5 - 2) * 2 + 5

3*2 + 5

6 + 5

= 11

2, 5, 11, 23, 47, 95, 191

2. A, Z, B, Y, C, ...

First letter , 26th letter, 2nd letter, 25th letter, 3rd letter, 24th letter, 4th letter, 23rd letter, 5th letter, 22nd letter

A, Z, B, Y, C, X, D, W, E, V

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What is the area of this figure?
Vladimir79 [104]

Answer:

<em>97 square miles</em>

Step-by-step explanation:

Let us divide this figure into rectangles, each of set dimensions;

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<em>Solution; 97 square miles</em>

4 0
3 years ago
Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

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<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%20%2B%5Cinfty%7D%20%28%5Cfrac%7Bx%5E%7B2%7D-3x-1%20%7D%7Bx%5E%7B2%7D%20-4x
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Step-by-step explanation:

I know but I m feeling bore

7 0
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Use the Quadratic formula, 

x= -b +/- sqrt(b^2-4ac)
     -------------------------
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x= 4 +/- sqrt(16+12)
     ----------------------
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x = 4 +/- 2*sqrt(7)
      ------------------
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     ------------------
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x= 2+/- sqrt(7)
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ki77a [65]
What is the full problem

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