I need help with these soon for the marking period

Question : 1/9X ⅝ + (½)³=

Flauer [41]

Answer:

Step-by-step explanation: 1/9X ⅝ + (½)³ = 0.19

8 0

3 years ago

Tesouo has 50 tockens each game at the aracde cost 4 tockens how many games can he play

Phoenix [80]

Answer: 12.5

Step-by-step explanation:

50 divided 4 equals 12.5 so he can play 12 games, but the math says the answer is 12.5, so you can choose between 12, and 12.5

5 0

3 years ago

Using the diagram on the right, find the length of GT and TA.

Art [367]

Answer:

$GT=10\ units$ and $TA=6\ units$

Step-by-step explanation:

we know that

In the diagram

Triangles GRT and GEA are similar by AA Similarity Postulate

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional

so

$\frac{GR}{GE}=\frac{GT}{GA}$

we have

$GR=5\ units$

$GE=5+3=8\ units$

$GT=16-x\ units$

$GA=16\ units$

substitute

$\frac{5}{8}=\frac{16-x}{16}\\ \\5*16=8*(16-x)\\ \\80=128-8x\\ \\8x=128-80\\ \\x=48/8\\ \\x=6\ units$

therefore

$GT=16-x=16-6=10\ units$

$TA=x=6\ units$

5 0

3 years ago

I need to solve for x

umka21 [38]

Answer:

dkdksdldlxlxzkzkdisi

6 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 ...

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 ...

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