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3 years ago

26. A positive whole number is called stable if at least one of its digits has the same value

Mathematics

1 answer:

Kisachek [45]3 years ago

5 0

Answer:

Step-by-step explanation:

Given the value 78247 a s a stable number because at least one of its digits has the same value as its position in the number. The 4th number in the value is 4, this makes the number a stable number.

The following are the 3-digits stable numbers that appears in 78247

The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).

Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.

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a (1,4) b (-1,2) c (5,-2) are the vertices of triangle abc find the sum of coordinates of the point where the right bisector of

zmey [24]

Answer:

Step-by-step explanation:

Find the midpoint of BC:

midpoint = (-1+5)/2, (2-2)/2 = (2, 0).

The slope of BC = (2 - -2) / (-1-5) = -2/3.

Find the equation of the right bisector of BC:

The slope = -1 / -2/3 = 3/2.

y-y1 = m(x-x1)

y - 0 = 3/2(x - 2)

y = 3/2x - 3.

Now find the equation of the median through C:

The midpoint of AB = (1 - 1)/2, (4+2)/2

= (0, 3).

The equation of the median:

The slope = (-2-3) / (5-0)

= -1.

The equation is:

y - 3 = -1(x - 0)

y -3 = -x.

Now we find the point of intersection by solving the 2 equations:

y - 3 = -x

y = 3/2x - 3

y = -x + 3

So:

3/2x - 3 = -x + 3

3/2x + x = 6

5/2 x = 6

x = 12/5.

y = -12/5 + 3

= -12/5 + 15/5

= 3/5.

The sum of the coordinates = 12/5 + 3 /5

= 15/5

= 3.

3 0

3 years ago

Suppose JK bisects angle LJM. The measure of angle LJK = (-10x + 3) and the measure of KJM = (-x + 21)º. Find the measure of ang

SpyIntel [72]

LJK = KJM -10x +3 = -x + 21 -10x + x = 21 - 3 -9x = 18 x = -2 KJM = -x + 21 = 2 + 21 = 23 LJM = LJK + KJM LJM = 23 + 23 LJM = 46º..

i think thats how its done

7 0

3 years ago

Which quiz did met have a better score?

MaRussiya [10]

I'm her first quiz she did better, this is going to be hard to explain but pay attention 92% as a decimal is 92/100 keep this in mind, then 27 out of 30 question we try to make 30 go into 100 but we know 30 can't go into 100 without a decimal so what we do is we change the decimal to 90 so for the 27 out of 30 it is now 81/90, now we need to change the denominator in 92/100 to 90 so we subtract 10 from the numerator and denominator to get 82/100 so compare the 2 decimals you got 81/90 and 82/90 and you find that 82/90 is bigger so she did better on the first quiz

7 0

3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

Given:

$y=\left(x+\sqrt{1+x^2}\right)^m$

First derivative

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

Second derivative

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

Proof

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

1 year ago

. A segment has a midpoint at (2, -7) and an endpoint at (8, -5). What are the coordinates of the other endpoint?

Sliva [168]

Ok, you can refer to the midpoint formula to find the endpoint. Here goes...

MP=(2,-7) and EP=(8,-5)
Let x represent the missing endpoint.
(8+x)/2=2 NOTE: =2 represents first number of MP and the representation of number 8 is self explanatory. You have two endpoints but need to identify the other endpoint so you divide by 2. Then, multiply by two on both sides.
2(8+x)/2 = 2*2
16+x/2=4 do the next step (simplify) on the left side of equation 16x/2=8
Now, subtract 4-8=-4 So, the x coordinated of the missing endpoint is -4.

4 0

3 years ago

Read 2 more answers