All categories

3 years ago

Hello

Mathematics

2 answers:

agasfer [191]3 years ago

4 0

hello

Step-by-step explanation:

Ayşe ben bertha

earnstyle [38]3 years ago

3 0

Answer:

Step-by-step explanation:

1 toy car is 3 so if you multiply it by 4 becuase theres 4 cars you get 12

he paid $12

he sells one car for $5 so multiply it by 4

and he would get $20 dollars wich is only $8 more dollars

You might be interested in

Can Someone Help Me With This?? :)

kykrilka [37]

Rational) No
Integer) No
Whole) Yes
Natural) Yes
Describe the number:
It’s whole and Natural and it is 5

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 ...

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

_____

<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 plane flying overhead at night is located by two

Gemiola [76]

Answer:

Step-by-step explanation:

Let the distance of two consecutive stones are x, x+1.

In ΔBCD, we have

tan60

⇒x=

.....(i)

In ΔABC, we have

tan30

x+1

⇒

x+1

⇒

+1=

h ......[from equation (i)]

⇒

⇒h=

solution

6 0

3 years ago

Given the point and the slope, write the function in point-slope form:<br> (2, -5); m = 2/3

pshichka [43]

Y=2/3x+(-5)
hope this helps

3 0

2 years ago

An English teacher has 6 short stories, 4 novels, and 23 poems to choose from. How many ways can he assign one of each to his cl

Karolina [17]

Answer:

552

Step-by-step explanation:

This is a problem of permutation which can be solved by rule of fundamental counting principle.

This principle states that if there "m" ways of doing one thing and "n" ways of doing other. Then no. of ways in which both the things can be done together is "m*n". This can be extended for m, n, p,r, s things and so on.

example: if there are 5 shirts and 3 trousers then number of ways in which the shirts and trousers can be worn is 5*3 = 15 ways.

_____________________________________________

The given problem is on similar concepts.

here 6 short stories, 4 novels, and 23 poems have to be assigned to his class.

Thus it can be done in 6*4*23 = 552 ways.

5 0

3 years ago