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3 years ago

Use the figure to find the measures of Z1 and 22

Mathematics

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

∠ 1 = 149°, ∠ 2 = 31°

Step-by-step explanation:

∠ 1 and 31° are same-side interior angles and are supplementary, thus

∠ 1 = 180° - 31° = 149°

∠ 2 and 31° are alternate angles and congruent, thus

∠ 2 = 31°

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True or False: As long as the information reported follows the generally accepted accounting principles (GAAP) guidelines, accou

olchik [2.2K]

Answer:

The given statement is true.

Step-by-step explanation:

Yes this is true.

GAAP is a collection of certain standard accounting rules for financial reporting.

Few general principles of GAAP guidelines are :

1. Principle of Regularity.

2. Principle of Sincerity.

3. Principle of Consistency.

4. Principle of Non-Compensation.

5. Principle of Continuity.

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

3 years ago

Q/19 < 4 pls be correct if you type it !!!!

babymother [125]

Answer:

q<76

Step-by-step explanation:

hope this is the answer you were looking for!

3 0

3 years ago

What is -15x=0 solution

cricket20 [7]

Here is the method for solving that equation:

Step #1: Write the equation:

-15x = 0

Step #2: Divide each side of the equation by -15 :

x = 0

3 0

3 years ago

write an equation in slope intercept form for a line through the following: through (5,2). with a slope of -2

Veseljchak [2.6K]

Answer:

y = - 2x + 12

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 2, thus

y = - 2x + c ← is the partial equation

To find c substitute (5, 2) into the partial equation

2 = - 10 + c ⇒ c = 2 + 10 = 12

y = - 2x + 12 ← equation of line

4 0

3 years ago