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3 years ago

Find the product of 3ab (10b + 2a)

Mathematics

1 answer:

fgiga [73]3 years ago

5 0

This would be the answer

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A $250 gaming system is discounted by 25%. It was then discounted another 10% because the box was opened by a customer. a. The s

iris [78.8K]

Answer:

a. Sale Price $=\$168.75$

b. Total percent discount $=32.5\%$

Step-by-step explanation:

First Discount:

Initial Price $=\$250$

Discount $=25\%$

$25\%\ of\ 250=\frac{25}{100}\times 250=62.5\\\\Now\ Price=250-62.5=$187.5$

Second Discount:

$Price=187.5\\\\Discount=10\%\\\\10\%\ of\ 187.5=\frac{10}{100}\times 187.5=18.75\\\\Now\ Sale\ Price=187.5-18.75=\$168.75$

Effective Discount:

Initial Price $=\$250$

Final Price $=\$168.75$

Total discount $=250-168.75=81.25$

$\%\ discount=\frac{81.25}{250}\times 100=32.5\%$

3 0

3 years ago

If 14y - 4 = 24y + 26 , then 12y =

harkovskaia [24]

Solutions

To solve the equation we have to isolate 12y

$14y - 4 = 24y + 26

-4 - 26 = 24y - 14y

-30 = 10y

10y = -30

y =- -30/10

y = -3

12y =12/-3

12y = -36

Therefore, 12y = -36$

4 0

3 years ago

Read 2 more answers

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 ...

$a[n]=(p+qn)r^n$