Ask question

Ask question

All categories

2 years ago

11

Above is a table that gives the interest per every $100 financed. Use the table to determine the annual percentage rate for a 11

month loan that charges $6.62 per every $100 financed.

1 answer:

Sladkaya [172]2 years ago

4 0

13 percent , just took the quiz

You might be interested in

Find the equation of the normal to the curve y=x^2-2x^2-4x+1 at thepoint (-1,2)

balandron [24]

Answer:

$y=-\frac{1}{3}x+\frac{5}{3}$

Step-by-step explanation:

Assuming that the first exponent in the formula for the curve should be 3, not 2...

$y=x^3-2x^2-4x+1$

The derivative is

$y'=3x^2-4x+4$

The slope of the tangent line at the point (-1, 2) is the value of the derivative at x = -1.

$y'|_{x=-1} = 3(-1)^2-4(-1)+4=3-4+4=3$

The slope of the normal line is the opposite reciprocal of the slope of the tangent line.

$m_{\text{normal} = -\frac{1}{3}$

Using the Point-Slope form of a linear equation, the normal line is

$y-2=-\frac{1}{3}(x-(-1))\\y-2=-\frac{1}{3}(x+1)\\y=-\frac{1}{3}x-\frac{1}{3}+2\\y=-\frac{1}{3}x+\frac{5}{3}$

5 0

2 years ago

What fraction represents 0.1875 ft?

aliya0001 [1]

Answer:

There is not much that can be done to figure out how to write 0.1875 as a fraction, except to literally use what the decimal portion of your number, the .1875 , means. Since there are 5 digits in 1875 , the very last digit is the "100000th" decimal place. So we can just say that .1875 is the same as 1875/100000.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What radian measure is equivalent to 900°?<br> 5n<br> 10n<br> 1/5n<br> 9n

Sergeu [11.5K]

9n is the answer to this question

3 0

2 years ago

Read 2 more answers

In a competitive examination negative marks of 2 are awarded for each wrong answers and 4 marks for every correct answer. If nee

Alex777 [14]

9514 1404 393

Answer:

Rajan gets 6 marks more

Step-by-step explanation:

Neerav's marks are (15)(4) +(5)(-2) = 60 -10 = 50.

Rajan's marks are (14)(4) = 56.

Rajan gets 6 more marks than Neerav.

3 0

2 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago