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3 years ago

Both fractions in the expression 1/3 + 1/3 are examples of ___________

Mathematics

2 answers:

Murljashka [212]3 years ago

8 0

2/6+2/6 just make them the same number adding

navik [9.2K]3 years ago

5 0

2/6 +2/6 or 4/12+4/12 and there are many more!

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Stan has made a $125.30 monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years. he would now

Andrew [12]

The annuity of the monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years is $333.71

<h3>What is annuity?</h3>

An annuity is a series of payments made at equal period of time.

future value = annuity x [(1 + i)ⁿ - 1] / i

annuity = $125.30

i = 1.5% / 12 = 0.00125

n = 35 years x 12 months = 420

future value = $125.30 x [(1 + 0.00125)⁴²⁰ - 1] / 0.00125

future value = $69,156.049 ≈ $69,156.05

annuity = [i x (present value)] / [1 - (1 + i)⁻ⁿ]

i = 1.5% / 12 = 0.00125

n = 20 years x 12 months = 240

present value = $69,156.05

annuity = (0.00125 x $69,156.05) / [1 - (1 + 0.00125)⁻²⁴⁰]

annuity = $86.45 / 0.25904

= $333.71

Learn more about annuity;

brainly.com/question/23554766

4 0

2 years ago

What is the approximate area of a sector given O= 56 degrees with a diameter of 12m?

jeyben [28]

Area of sector is 17.584 meters

<em><u>Solution:</u></em>

Given that we have to find the approximate area of a sector given O= 56 degrees with a diameter of 12m

Diameter = 12 m

Radius = Diameter / 2 = 6 m

An angle of 56 degrees is the fraction $\frac{56}{360}$ of the whole rotation

A sector of a circle with a sector angle of 56 degrees is therefore also the fraction $\frac{56}{360}$ of the circle

The area of the sector will therefore also be $\frac{56}{360}$ of the area

$\text{ sector area } = \frac{56}{360} \times \pi r^2\\\\\text{ sector area } = \frac{56}{360} \times 3.14 \times 6^2\\\\\text{ sector area } = 17.584$

Thus area of sector is 17.584 meters

3 0

3 years ago

PLZ HELP ME I AM TIMED HELP PLZ

ladessa [460]

Answer:

ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

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 ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

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

What is the area of a circle that has a diameter of 18"?

lubasha [3.4K]

Since the formula for the area of a circle is Pi x r squared, the area would be 81 x Pi (you teacher may want you to use Pi as 3.14 or put it in your calculator).

7 0

3 years ago

Read 2 more answers