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choli [55]
3 years ago
14

When constructing an inscribed square, how many lines will be drawn in the circle? (6 points) A. 2 B. 3 C .5 D. 7

Mathematics
1 answer:
nataly862011 [7]3 years ago
4 0

The answer is 2 or A.

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If using the method of completing the square to solve the quadratic equation
Makovka662 [10]

Answer:

=-1+\sqrt{15}i,\:x=-1-\sqrt{15}i

Step-by-step explanation:

7 0
3 years ago
Is it possible for a 5×5 matrix to be invertible when its columns do not span set of real numbers R^5​? Why or why​ not? Select
xeze [42]

Answer:

The correct answer is A..

Step-by-step explanation:

From the Invertible Matrix Theorem (IMT) we have a set of equivalent conditions to determine if a square matrix is invertible or not. In particular, it says that a square matrix of dimension tex]n\times n[/tex] is invertible if and only if, its columns span the vector space tex]R^n[/tex].

In the particular case of this exercise we have a matrix of dimension tex]5\times 5[/tex]. So, by the Invertible Matrix Theorem its columns must span the vector space tex]R^5[/tex]. Now, according to the statement of the exercise this condition does not hold. Hence, the given matrix cannot be invertible.

5 0
3 years ago
What are the factors of -3?
elena55 [62]

Step-by-step explanation:

The positive factors of -3 are 3 and 1

There is also -3 and -1

6 0
3 years ago
What is the solution set of the equation using the quadratic formula? x2+6x+10=0 {−3+2i,−3−2i} {−6+2i,−6−2i} {−3+i,−3−i} {−2i,−4
olga2289 [7]

Answer:

The solution set of the quadratic function x^{2}+6\cdot x +10 is \{-3+i,-3-i\} .

Step-by-step explanation:

Let be a second-order polynomial (quadratic function) is standard form and equalized to zero:

a\cdot x^{2}+b\cdot x + c = 0

Its roots can be determined by the Quadratic Formula in terms of its polynomial coefficients, which states that:

x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

Given that a = 1, b = 6 and c = 10, the roots of the polynomial are, respectively:

x_{1,2} = \frac{-6\pm \sqrt{6^{2}-4\cdot (1)\cdot (10)}}{2\cdot (1)}

x_{1,2} = -3\pm i

x_{1} = -3+i

x_{2} = -3 -i

The solution set of the quadratic function x^{2}+6\cdot x +10 is \{-3+i,-3-i\} .

8 0
3 years ago
How many 9 are present from 1 to 10000​
astraxan [27]

Answer:

300

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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