Solve for x A) -5 B) 6 C) -9 D) 1

Look at attachment 10 POINTS!!!

Leni [432]

Answer:

The midpoint of TS is (-1,-3)

The coordinates of M should be (8,18)

Step-by-step explanation:

3 0

3 years ago

The next model of a sport car will cost 3.6% more than the current model. The current model cost $42,000. How much will the pric

Murrr4er [49]

Answer:

$43,512

Step-by-step explanation:

3.6% = 0.036

0.036 * 42,000 = 1,512

42,000 + 1,512 = 43,512

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 ...

a)

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

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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 ...

d)

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

__

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

f)

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

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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

3 years ago

What is the solution to this system of equations?

Maurinko [17]

Answer:

(x, y) = (5, -2)

Step-by-step explanation:

A graphing calculator provides a quick and easy way to find the solution.

_____

There are several other ways to solve these equations. Or you can estimate where the answer might be using logic like this:

The intercepts of the first equation are ...

x-intercept = 26/4 = 6 1/2
y-intercept = -26/3 = -8 2/3

So the graph of it will form a triangle with the axes in the 4th quadrant.

The intercepts of the second equation are ...

x-intercept = 11/3 = 3 2/3
y-intercept = 11/2 = 5 1/2

So the graph of it will form a triangle with the axes in the 1st quadrant. The x-intercept of this one is less than the x-intercept of the first equation, so the two lines must cross in the 4th quadrant.

The only 4th-quadrant answer choice is (5, -2).

3 0

3 years ago

12 ten thousands, 8 thousands,14 hundreds, 7 ones in standard form

coldgirl [10]

12 x 10000 = 120000
8 x 1000 = 8000
14 x 100 = 1400
7 x 1 = 7

120000 + 8000 + 1400 + 7
129407 is your answer

hope this helps

8 0

3 years ago