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VLD [36.1K]
3 years ago
15

Aphrodite took out a 30-year loan from her bank for $170,000 at an APR of

Mathematics
1 answer:
horsena [70]3 years ago
8 0

Answer:

  A. $3246.74

Step-by-step explanation:

The monthly payment can be found from the amortization formula.

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the principal amount, r is the annual rate compounded n times per year for t years.

Filling in the values, we compute the monthly payment to be ...

  A = $170,000(.072/12)/(1 -(1 +.072/12)^(-12·30)) = $1153.94

__

The remaining balance after t years will be ...

  B = P(1 +r/n)^(nt) -A((1 +r/n)^(nt) -1)/(r/n)

For the given initial principal and the computed payment, after 18 years, the balance will be ...

  B = $170000(1 +.072/12)^(12·18) -$1153.94((1 +.072/12)^(12·18) -1)/(.072/12)

  B = $111,054.71

The prepayment penalty appears to be ...

  (r/2)(0.80B) = (.072/2)(0.80)($111,054.71) = $3,198.38

The closest listed answer choice is ...

  A.  $3246.74

_____

Please ask your teacher how to get the answer, since none of the offered choices appear to be correct.

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Fined the slope of this table
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Answer:

−9

Step-by-step explanation:

\frac{y2-y1}{x2-x1} \\\\= \frac{0-49}{0-(-5)} \\\\= \frac{-45}{5} \\= -9

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FIND THE GCF. Mathematics
Rom4ik [11]

The G.C.F of the given algebraic expression is; ⁴⁹/₆a xy

<h3>What is the G.C.F (Greatest Common Factor)?</h3>

The greatest common factor (GCF or GCD or HCF) of a set of whole numbers is the largest positive integer that divides evenly into all numbers with zero remainder. For example, for the set of numbers 18, 30 and 42 the GCF = 6.

We are given the algebraic expression;

(3 x y * 4x²y * ¹/₁ * 5x³y ^ z * 4)7a * 1/1 * 7*6

Expanding this further gives;

⁴⁹/₆a((3xy * 4x²y * 20x³y ^ z )

Now, the GCF of the terms inside the bracket would be x y. Thus, expanding the GCF  we have;

⁴⁹/₆a xy((3 * 4x * 20x²y^(z - 1))

Read more about G.C.F at; brainly.com/question/219464

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5 0
1 year ago
Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0
andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0   ==>   5<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   5<em>x</em> = 2<em>nπ</em>   <u>or</u>   5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = 2<em>nπ</em>/5   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0   ==>   2<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   2<em>x</em> = 2<em>nπ</em>   <u>or</u>   2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = <em>nπ</em>   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2   ==>   2<em>x</em> = arccos(-1/2) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … …    ==>   2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0
2 years ago
We can describe 15×-10 as an expression. we would describe 6×-2&lt; 35 as an...
algol [13]

Answer: Inequality

Step-by-step explanation:

From the question, if we can describe 15×-10 as an expression, then we would describe 6×-2< 35 as an inequality. An

inequality is used to compare two values, and shows if one is less than, or greater than, or maybe not equal to the other value.

For example, a ≠ b means that a is not equal to b and a < b means a is less than b while a > b means a is greater than b. From the question, 6×-2< 35 means that 6x - 2 is less than 35.

8 0
3 years ago
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