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iren2701 [21]
2 years ago
9

the following sequence is an arithmetic sequence.you are now shown very many terms of this sequence,but you are shown enough to

fill in the missing number.what number goes in the blank? 14, _,56 ...?
Mathematics
1 answer:
aivan3 [116]2 years ago
8 0

because it is arithmetic not geometric 14,x,56 An = An-1 + d x = 14 +d --> d = x-14 56 = x +d --> d = 56-x By substitution x-14 = 56-x 2x = 70 x = 35 the sequence is 14,35,56

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As 0.1333 is less than 0.5 the answer is


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Gnom [1K]

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Find the number of permutations of the first 12 letters taken 3 at a time.
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7 0
2 years ago
A gun mass of 5 kg fired a bullet of mass 10 g with the velocity of 360 km/h. What
klemol [59]

Answer:

The recoil velocity of the gun is 0.72\,\,\frac{km}{h}  and is pointing in opposite direction to the velocity of the bullet.

Step-by-step explanation:

Use conservation of linear momentum, which states that the momentum of the bullet (product of the bullet's mass times its speed) should equal in absolute value the momentum of the recoiling gun (its mass times its recoil velocity).

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In mathematical terms, we have:

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6 0
3 years ago
Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

rerange
\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
2 years ago
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