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4 years ago

Please help asap ! What is the area of a circle with a radius of 6 inches?

Mathematics

2 answers:

Gwar [14]4 years ago

7 0

Answer:

36pi in²

Step-by-step explanation:

Area = pi × r²

= pi × 6²

= 36pi

jeyben [28]4 years ago

4 0

Answer:

A = 36 pi in^2

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

A = pi (6)^2

A = 36 pi in^2

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please help! functions and relations. f(x)= square root of x-4. find the inverse of f(x) and it’s domain.

likoan [24]

ANSWER:

$\:D.\text{ }f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge-4$

STEP-BY-STEP EXPLANATION:

We have the following equation:

$f(x)=\sqrt{x}-4$

The inverse is the following (we calculate it by replacing f(x) by x and x by f(x)):

$\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\ \\ \sqrt{f^{-1}(x)}=x+4 \\ \\ f^{-1}(x)=(x+4)^2 \end{gathered}$

The domain would be the range of the original equation, and it would be the range of values that f(x) could take, which was from -4 to positive infinity, that is, f(x) ≥ -4.

Therefore, the domain is x ≥ -4.

So the correct answer is D.

$\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4$

4 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 ...

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

The initial condition equations are now ...

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

and the solutions for p and q are ...

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

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

Suppose ABCD is a rhombus with AB = 12 inches. The midpoints of its sides are joined to form a quadrilateral. What type of quadr

MrRissso [65]

Answer:

A rectangle

Step-by-step explanation:

The given parameters are;

The quadrilateral ABCD is a rhombus;

The length of the sides of the rhombus = 12 inches

The lengths of the sides of a rhombus are equal

The opposite interior angles of the rhombus are equal

The length of each midpoint from the vertex = 12 in./2 = 6 in.

Therefore, we have;

The line joining the midpoints form a quadrilateral with the length of the opposite sides equal

The sum of the interior angles of the rhombus = 180°

From the diagram created with Microsoft Visio, we have;

4·a + 4·b + 360 = 4 × 180 = 360 + 360

4·a + 4·b = 360

a + b = 90°

We have;

The interior angles of the quadrilateral formed = x

a + b + x = 180° Sum of angles on a straight line

∴ x = 180° - (a + b) = 180° - 90° = 90°

x = 90°

Therefore, the interior angles of the quadrilateral formed = 90°

The quadrilateral formed having equal opposite sides, and all interior angle of 90° each is a rectangle

6 0

3 years ago

can somebody please write what are the rules for multiplying integers and the rules for dividing integers in a short answer?

arsen [322]

Answer:

Negative + Negative = Positive

Negative + Positive = Negative

Positive + Negative = Negative

Positive + Positive = Negative

The rules are the same for division.

Step-by-step explanation:

4 0

3 years ago

What is the mode of the following set of numbers: 10, 20, 30, 20, 30, 10, 30, 20, 30, 15 *

likoan [24]

Answer:

A. 30

Step-by-step explanation:

Because it showed up the most

6 0

3 years ago

Read 2 more answers