Which of the following are measurements of the sides of a right triangle?

An actor invests some money at 9% simple interest, and $24,000 more than three times the amount at 10% simple interest. The

Andrew [12]

Answer:

The amount invested at 9% is $93000

The amount invested at 10% is $303000

Step-by-step explanation:

Let the amount invested at 9% interest rate be x

And the amount invested at 10% rate be y

Simple Interest from x in a year = 0.09x

Simple Interest from y in a year = 0.1y

But y = 24000 + 3x

And the sun of the interests, 0.09x + 0.1y = 38670

Now we have a simultaneous eqn

y = 24000 + 3x (eqn 1)

0.09x + 0.1y = 38670 (eqn)

Substitute y into eqn 2

0.09x + 0.1(24000 + 3x) = 38670

0.09x + 2400 + 0.3x = 38670

0.39x = 38670 - 2400

x = 36270/0.39 = $93000

y = 24000 + 3x = 24000 + 3 × 93000 = $303000

8 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

3 years ago

An oblique prism with a square base of edge length x United has a volume of 1/2 x3 cubic units. Which expression represents the

aleksklad [387]

Answer:

The answer is x units.

Step-by-step explanation:

We are tasked to solve for the expression that represents the height of the prism. We are given with the following values:

length = x² units

volume = x³ units

We will use the formula below:

V = length x height

x³ = x² * height

height = x³/x units

height = x units

The answer is x units.

7 0

2 years ago

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What is the height of an equilateral triangle with side lengths of 15? round to tenths?

Vinil7 [7]

If this is an equilateral triangle, we need half of it to find the height which is a leg of a right triangle with a hypotenuse of 15. That means that the base of this half triangle is 7.5. Use Pythagorean's Theorem to fine the height. $x^{2} =(15) ^{2} -(7.5) ^{2}$ . x = 12.99 or 13

8 0

3 years ago

3/4 x 1/2 <br> Help need it plss ;-;

Shkiper50 [21]

Answer:

3/8

Step-by-step explanation:

3/4 divide 2 = 4÷2= 3/8

4 0

3 years ago

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