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3 years ago

Diagram 1 shows a composite solid consisting of a cuboid and a right pyramid.

Mathematics

1 answer:

kondaur [170]3 years ago

4 0

44 the total volume of this composite solid is 896

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12 divided by 1 1/5 = ? IN lowest TERM

kompoz [17]

Answer:

12 divided by 1 1/5 = 60/11 in lowest term

I hope this helps you.

Step-by-step explanation:

Simplify the expression.

5 0

3 years 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 ...

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

_____

Comment on problem g

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

6 0

2 years ago

Karl is building a rectangular garden bed. The length is 6 feet. She has 20 feet of boards to make the sides. Write and solve an

natali 33 [55]

Answer:

The width of the garden bed must be less than or equal to 4 feet.

$w\leq4$

Explanation:

Given that;

She has 20 feet of boards to make the sides.

The perimeter of the garden bed must not be more than 20 feet

$\begin{gathered} P=2l+2w\leq20 \\ 2l+2w\leq20 \end{gathered}$

Given;

The length is 6 feet;

$l=6$

To get the inequality for the width w, let us substitute the value of the length into the inequality above and simplify;

$\begin{gathered} 2l+2w\leq20 \\ 2(6)+2w\leq20 \\ 12+2w\leq20 \\ 2w\leq20-12 \\ 2w\leq8 \\ \frac{2w}{2}\leq\frac{8}{2} \\ w\leq4 \end{gathered}$

Therefore, the width of the garden bed must be less than or equal to 4 feet.

$w\leq4$

8 0

1 year ago

Find the domain of the graphed function?

MariettaO [177]

Remember, the domain of a function is all of the x values in the (x,y) coordinates. For this, let’s take a few bench mark coordinates. (0,0) (1,2.5)(2,0) (3, -9). Those definitely aren’t all real numbers, so eliminate A. X is not only less than or equal to 0, so eliminate C.

It’s B.

7 0

3 years ago

Read 2 more answers

Will someone please answer these questions fast and I will give the brainliest? I really need help and it would be appreciated.

Anika [276]

Hello there,

(1) $7934.37

interest would be $2934.37

(2) $2593.74

interest would be $1593.74

(3) $1469.33

interest would be $469.33

(4) $1296.87

interest would be $796.87

(5) $239.66

interest would be $139.66

(6) $4660.96

interest would be $3660.96

Hope I Helped!

-Char

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2 years ago