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Ganezh [65]
3 years ago
12

Barbara has $70.00 in her pocket and a $20 gift card. which purchase will she be able to make without having to use her credit c

ard?
Mathematics
2 answers:
Tresset [83]3 years ago
4 0

Answer:

Step-by-step explanation:

six puzzles for $11.99 each and a board game for $15.89

Kipish [7]3 years ago
3 0
If Barbara has $ 70.00 in her pocket and a $ 20 gift card. then the maximumm amount she can pruchase is $ 70 + $ 20 = $ 90
so let x be purchase will she be able to make without having to use her credit card
x >= $ 90
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Can you help me i Don’t understand
snow_tiger [21]

Answer:

  A.  {-1, 2, 7}

Step-by-step explanation:

For a set of ordered pairs, the domain is the set of first-numbers (x). The range is the set of second numbers (y, or g(x)).

The first step here is to find the g(x) values for each of the listed x-values.

The attachment shows a table.

Often, we like to list the domain and/or range values in numerical order, eliminating any duplicates. That is, the set of range values {2, -1, -1, 7} would be listed as ...

  {-1, 2, 7} . . . . . . matches choice A

_____

You find the g(x) values by substituting for x and doing the arithmetic.

  g(-2) = (-2)² -2 = 4 -2 = 2, for example

7 0
2 years ago
a five inch bamboo shoot doubles in height every three days. If the equation y=ab to the x power, where x represents the number
PtichkaEL [24]
This question was once asked before so I will put it here.

The correct answer for the question that is being presented above is this one: "x = 2; a = 5; b = 3" A 5 inch tall bamboo shoot doubles in height every 3 days. If the equation y=ab^x, where x is the number of doubling periods, represents the height of the bamboo shoot.

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

__

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
2 years ago
Yeah I need help…..<br> Do your work soldiers
Aloiza [94]

I can’t, sir! I cannot leave my post to help with your math problems, over!

8 0
2 years ago
Read 2 more answers
Find the length of base PQ in the trapezoid.
iris [78.8K]

Answer:

PQ = 20

Step-by-step explanation:

Theorem:  The length of the median of a trapezoid is the average of the base lengths.

\frac{PQ + 12}{2}  = 16

PQ + 12 = 32

PQ = 20

See how easy that is if you know the theorem!

5 0
3 years ago
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