Can someone help me?

Jacey obtains a 30-year 6/2 ARM at 4% with a 2/6 cap structure in the amount of $224,500. What is the monthly payment during the

beks73 [17]

In this case we have an ARM fixed for 6 years and adjust after the initial first 6 years every 2 years after. The basic idea behind a ARM is that the interest changes periodically, but since our ARM is fixed for 6 years, our going to calculate the monthly payment during the initial period using the formula: $m= \frac{P( \frac{r}{12}) }{1-(1+ \frac{r}{12})^{-12t} }$
where
$m$ is the monthly payment
$P$ is the amount
$r$ is the interest rate in decimal form
$t$ is the number years

First we need to convert our interest rate of 4% to decimal form by dividing it by 100%:
$\frac{4}{100} =0.04$
We also know from our question that $P=224500$ and $t=30$ , so lets replace those values into our formula to find the monthly payment:
$m= \frac{224500( \frac{0.04}{12}) }{1-(1+ \frac{0.04}{12})^{-(12)(30)} }$
$m=1071.80$

We can conclude that the monthly payment during the initial period is $1071.58<span />

7 0

3 years ago

Find the missing number. show working. 13+?=15+1

saw5 [17]

13+2=15+1=16 that is the answer

7 0

3 years ago

Read 2 more answers

If ABCD is a parallelogram, what can we say about the diagonals BD and AC?

lidiya [134]

B. The diagonals bisect each other.

4 0

3 years ago

Read 2 more answers

g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and

allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, $[ab]_K$ is equal to $[ba]_K$ , thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, $[ab]_H = [ba]_H$ . Since a and b were generic elements of H, then H/G is abelian.

4 0

3 years ago

Select the two values of x that are roots of this equation.<br> 2x^2+ 1 = 5x

iren [92.7K]

Answer:

$\frac{5+\sqrt{17}}{4}$ , $\frac{5-\sqrt{17}}{4}$

Step-by-step explanation:

One is asked to find the root of the following equation:

$2x^2+1=5x$

Manipulate the equation such that it conforms to the standard form of a quadratic equation. The standard quadratic equation in the general format is as follows:

$ax^2+bx+c=0$

Change the given equation using inverse operations,

$2x^2+1=5x$

$2x^2-5x+1=0$

The quadratic formula is a method that can be used to find the roots of a quadratic equation. Graphically speaking, the roots of a quadratic equation are where the graph of the quadratic equation intersects the x-axis. The quadratic formula uses the coefficients of the terms in the quadratic equation to find the values at which the graph of the equation intersects the x-axis. The quadratic formula, in the general format, is as follows:

$\frac{-b(+-)\sqrt{b^2-4ac}}{2a}$

Please note that the terms used in the general equation of the quadratic formula correspond to the coefficients of the terms in the general format of the quadratic equation. Substitute the coefficients of the terms in the given problem into the quadratic formula,

$\frac{-b(+-)\sqrt{b^2-4ac}}{2a}$

$\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}$

Simplify,

$\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}$

$\frac{5(+-)\sqrt{25-8}}{4}$

$\frac{5(+-)\sqrt{17}}{4}$

Rewrite,

$\frac{5(+-)\sqrt{17}}{4}$

$\frac{5+\sqrt{17}}{4}$ , $\frac{5-\sqrt{17}}{4}$

8 0

3 years ago