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bixtya [17]
3 years ago
7

The sum of the digits of a two-digit number is 7. If the digits are reverses, the new number is 9 more than the original number.

Find the original number. ​
Mathematics
1 answer:
____ [38]3 years ago
5 0

25

Step-by-step explanation

Let the ones's digit =x.

So, sum of the digits =7

Tens's digit =7−x

Original no. =10(7−x)+x=70−10x+x=70−9x

Reversed no. =10x+7−x=9x+7

2(orig no) = reversed no. −2

2(70−9x)=9x+7−2

140−18x=9x+5

27x=135

x=5

Ten's digit =7−5=2

No. =25

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Perform the indicated operation. Simplify the result in factored form. ((x - y)/ (x^2 - 1)) * ((x - 1)/(x^2 - y^2))
Scilla [17]

Answer:

The answer to your question is            \frac{1}{(x + 1)(x + y)}            

Step-by-step explanation:

Real operation

                      \frac{x - y}{x^{2} - 1} \frac{x - 1}{x^{2} - y^{2}}

Process

1.- Factor the denominators

                     \frac{x - y}{(x + 1)(x - 1)} \frac{x - 1}{(x + y)(x - y)}

2.- Simplify like terms

Cancel (x - y) and (x - 1) because they are in numerator and the denominator

                    \frac{1}{(x + 1)(x + y)}            This is the answer or you can expand

3.- Expand

                   \frac{1}{x^{2} + xy + x + y^{2}}                    

3 0
3 years ago
Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect
Tasya [4]
4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1

defines an ellipse centered at (0,0) with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the x-axis. So if cross sections are taken perpendicular to the x-axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at x=x_0 will have a base of length

\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}
\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a \sqrt2:1 ratio with either of the remaining legs. So if the hypotenuse is \dfrac43\sqrt{9-{x_0}^2}, then either leg will have length \dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}.

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is \dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}.

Finally, x_0 can be chosen from any value in -3\le x_0\le3. We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)

and so the volume would be

\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx
=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}
=16

6 0
3 years ago
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