Answer:
The answer to your question is 0 < x ≤ 3/14
Step-by-step explanation:
Here are the steps to solve it.
First, you need to solve your inequality step-by-step:
3/x −2 ≥ 12
-2x + 3/x = ≥12
Next, find the critical points of the inequality:
2x + 3/x = 12
−2x + 3 = 12x (Multiply both sides by x)
−2x + 3 − 12x = 12x − 12x (Subtract 12x from both sides)
−14x + 3 = 0
−14x + 3 − 3 = 0 − 3 (Subtract 3 from both sides)
−14x = −3
-14x/ -14 = -3/ -14 (Divide both sides by -14)
So, that gives you your answer:
x = 3/14
Answer:
Step-by-step explanation:
Ive already explained this somebody else asked if you find it you will have the answer
The given circle having two external tangents, according to circle
theorem, have the following values;
<h3>How can the radial length and angles be found?</h3>
r² + 1.6² = (0.4 + r)²
(0.4 + r)² - (r² + 1.6²) = 0
0.8·r - 2.4 = 0
0.8·r = 2.4
r = 2.4 ÷ 0.8 = 3
y = 49°
From the two tangent angle theorem, we have;
m∠1 = ((82 + 98 + 82) - (360 - ((82 + 98 + 82)))) ÷ 2 = 82
According to circle theorems, angle at the center is twice angle
subtended at the circumference.
m∠2 = (360 - (82 + 98 + 82)) ÷ 2 = 49°
The length of the chord subtended by the arc <em>x</em> is equal to the length of
the chord subtended by 89°, which gives;
Learn more about circle theorems applications here:
brainly.com/question/16879446
Option D. D has the matrix of constants [[12], [11], [4]].
Step-by-step explanation:
Step 1:
With the given equations, we can form matrices to represent them.
The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.
Step 2:
The linear system A is represented as
![\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%261%261%5C%5C1%26-11%260%5C%5C1%26-1%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D26%5C%5C17%5C%5C23%5Cend%7Barray%7D%5Cright%5D)
.
Step 3:
The linear system B is represented as
![\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%261%261%5C%5C1%26-11%260%5C%5C1%26-1%2612%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D23%5C%5C17%5C%5C26%5Cend%7Barray%7D%5Cright%5D)
.
Step 4:
The linear system C is represented as
![\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C1%26-1%260%5C%5C1%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C11%5C%5C12%5Cend%7Barray%7D%5Cright%5D)
.
Step 5:
The linear system D is represented as
![= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%5C%5C11%5C%5C4%5Cend%7Barray%7D%5Cright%5D)
.
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
