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BlackZzzverrR [31]
3 years ago
15

Business Loss A company loses $40 as a result of a shipping delay. The 2 owners of the company

Mathematics
1 answer:
frozen [14]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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What’s the measure of str
IrinaVladis [17]

Answer:

I believe its 45 degrees.

5 0
3 years ago
-2x + y = 4<br> y = x + 2<br><br><br> Third question^ solve the system of equations
Leya [2.2K]

Answer:

x=-2 y=0

Step-by-step explanation:

put the second equation in to the first one

-2x+x+2=4

-x=2

x=-2

y=-2+2=0

x=-2 y=0

6 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

__

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

_____

<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
A swimming class has 20 students. Only 8 students know how to dive off
Allisa [31]

Answer: 40%  

Step-by-step explanation:

To write 8/20 as a percent have to remember that 1 equal 100% and that what you need to do is just to multiply the number by 100 and add at the end symbol % . 8/20 * 100 = 0.4 * 100 = 40% And finally we have: 8/20 as a percent equals 40%

6 0
3 years ago
Which shapes can be drawn with two opposite angles equal to 105°
sertanlavr [38]

Answer:

see the explanation

Step-by-step explanation:

we know that

A shape with two opposite angles equal to 105° could be a quadrilateral, a parallelogram, a rhombus or a trapezoid

Because

<em>A quadrilateral</em>: A quadrilateral is a four-sided polygon. The sum of the interior angles in any quadrilateral must be equal to 360 degrees

so

If the quadrilateral have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to

360^o-2(105^o)=150^o

<em>A parallelogram</em>: A Parallelogram is a flat shape with opposite sides parallel and equal in length. Opposite angles are congruent and consecutive angles are supplementary

so

If the parallelogram have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to

180^o-105^o=75^o

<em>A rhombus</em>: A Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond. All sides have equal length. Opposite sides are parallel. Opposite angles are congruent and consecutive angles are supplementary

so

If the Rhombus have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to

180^o-105^o=75^o

<em>A trapezoid</em>: A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel

so

If the trapezoid have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to

360^o-2(105^o)=150^o

4 0
3 years ago
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