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3 years ago

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Peter kim wanted to buy a new car.To help finance the purchase he decided to sell his organic markets bond in the secondary mark

et.Peters bond had a par value of $ 10,000 and a coupon of 6 percent.Current interests were 3 percent.What would peters bond sell for?

1 answer:

amid [387]3 years ago

3 0

At a premium, since it pays 6% and market is only 3%.

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4 years ago

Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

<u>Step 1</u>

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

<u>Step 2</u>

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

<u>Step 3</u>

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

<u>Step 4</u>

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

<u>Step 5</u>

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

<u>Step 6</u>

<u>Segment EF, segment FG, segment GH, and segment EH are congruent. </u>

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

<u>Step 7</u>

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

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Answer:

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