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pishuonlain [190]
3 years ago
8

I am an even number i am less than 100,the sum of my digits is 15, i am multiple of 12

Mathematics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer:

96

Step-by-step explanation:

herp derp herp derp herp derp

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ANSWERED
kvv77 [185]

Answer:

c but you have it so thanks

Step-by-step explanation:

6 0
3 years ago
Solve the problem.
Rufina [12.5K]
Answer is the B Sure let’s gooo
3 0
3 years ago
A state vector X for a four-state Markov chain is such that the system is four times as likely to be in state 2 as in 4, is not
GREYUIT [131]
All the components in the state vector need to sum to 1. You're given that component corresponding to state 1 is 0.2, and that the component for state 3 is 0.

That leaves states 2 and 4, for which you're told that the component for state 2 is four times as large. If x_i is the component for state i, then you have

x_1+x_2+x_3+x_4=1\iff 0.2+4x_4+0+x_4=1\implies5x_4=0.8\implies x_4=0.16

which means x_2=4x_4=0.64. So the state vector is \mathbf x=(0.2,0.64,0,0.16).
8 0
3 years ago
find the equation of the circle where (-9,4),(-2,5),(-8,-3),(-1,-2) are the vertices of an inscribed square.
solniwko [45]
Check the picture below, so, that'd be the square inscribed in the circle.

so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2}
\\\\\\
d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2}
\\\\\\
d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10

that means the radius r = 5.

now, what's the center?  well, the Midpoint of the diagonals, is really the center of the circle, let's check,

\bf \textit{middle point of 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-8-2}{2}~,~\cfrac{-3+5}{2} \right)\implies (-5~,~1)

so, now we know the center coordinates and the radius, let's plug them in,

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}\\
&-5&1&5
\end{array}
\\\\\\\
[x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25

8 0
3 years ago
Can someone please help me?
SpyIntel [72]

Answer:if you looked at hem like they were rows, in the first row the third one would be an expression. In the second row, the second one would be an expression. This is becuase they both  have an equal sign.

8 0
3 years ago
Read 2 more answers
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