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Citrus2011 [14]
4 years ago
10

Make a conjecture about the next item in the sequence. 1, 4, 16, 64, 256

Mathematics
2 answers:
kirza4 [7]4 years ago
8 0
1,4,16,64,256,1024 is the next number
True [87]4 years ago
3 0
1024, 4096, basically multiplying your previous number by 4
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Can someone please please help me!!!!!
Licemer1 [7]

Answer:

<h2>Diameter = 10</h2>

Step-by-step explanation:

If the side of a triangle inscribed in a circle is a diameter, then it is a right triangle.

Use the Pythagorean theorem:

d - diameter

d^2=8^2+6^2\\\\d^2=64+36\\\\d^2=100\to d=\sqrt{100}\\\\d=10

6 0
3 years ago
Help please im stuck
serg [7]
If the two equations have the same solution, they will have the same x value that makes them true. So solve for x for both equations and see if they are equal:

8x=87 \\&#10;x=10.875

\frac{2}{3} x+ \frac{3}{4} =8 \\&#10; \frac{2}{3} x=7.25 \\&#10;x=10.875

Because both equations have the same x value, they have the same solution.


7 0
4 years ago
Read 2 more answers
What is the equation for this graph
Nina [5.8K]

Answer: It’s y= 1/3x + 4

Step-by-step explanation: Because if you try to graph it then you get the points right. 4 will be on the y-axis of the chart.

5 0
4 years ago
How do you simplify a fraction?
attashe74 [19]
You simplify a fraction by finding it's greatest common factor. A greatest common factor is what both numbers are divisible by. If they can both be divided by two, then you've simplified them a little bit. Keep going until you can't find a common number that can be divided by the two, and you've got your simplified fraction. For example, 10/5 = 1/2 since both 10 and 5 is divisible by 5.
5 0
4 years ago
Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.
solong [7]

Answer:

A)\,\,det(A)=1

B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

Step-by-step explanation:

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|

Expanding with first row

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1

To find inverse we first find cofactor matrix

C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\

C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\

Cofactor matrix is

C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

4 0
3 years ago
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