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2 years ago

What is the square root

Mathematics

2 answers:

Hitman42 [59]2 years ago

8 0

Answer:

the square root is 2.2

FrozenT [24]2 years ago

3 0

Answer:

2.2

Step-by-step explanation:

make me brainestly

the guy who did the second answer wants beef

You might be interested in

Is 24 the correct LCD to use for subtracting 6 3/4 - 1 2/6 ? Why or why not?

zimovet [89]

Answer:

Step-by-step explanation:

No 24 is not the best number to use for the LCD, although it will work. 12 is much better because it is smaller.

Factor each number into it's primes.

24 =<u> 2*2*2</u>*3

32 = <u>2*2*2</u>*2*2

The highest common factor is 2*2*2 = 8

9/10 - 3/10 = 6/10 The denominator remains the same. The numerators get subtracted.

6/10 = 3/5 Two goes into both numerator and denominator of the answer. Therefore you can divide both 6 an 10 by 2

3 0

2 years ago

PLEASE HELP ASAP!!

natita [175]

We can infer and logically deduce that the arc intersected by these chords are not congruent.

<h3>What is a circle?</h3>

A circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Also, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Geometry, a circle is generally considered to be a conic section which is formed by a plane intersecting a double-napped cone that is perpendicular to a fixed point (central axis) because it forms an angle of 90° with the central axis.

In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.

Let the radii of the two (2) different circles with center O and O' be r₁ and r₂ respectively. Also, two (2) isosceles triangles AOB and COD would be formed and as such we have:

AO = OB = r₁.

CO = O'D = r₂.

By critically observing the diagram (see attachment) of these circles, we can logically deduce that the arc intersected by chord AB is greater than the arc intersected by the chord CD.

This ultimately implies that, the arcs intersected by these chords are not congruent.

Read more on arcs here: brainly.com/question/11126174

#SPJ1

<u>Complete Question:</u>

Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chords also congruent? Explain.

Hint: It would be helpful to draw two circles and label them according to the given information, then evaluate possible arc measures.

Consider the type of triangle that may be drawn by connecting the endpoints of a chord to the center of a circle. Compare the triangles made by two circles with different radii.

3 0

1 year ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$