Jaime needs a new roof on her house. The cash cost is $7500. She decides to finance the project by paying 20% down , with the balance paid in 36 monthly payments of $189.40. (a) What finance charge will Jaime pay? (b) What is the APR to the nearest half percent ?

Answers

Answer 1

Answer:

(a) What finance charge will Jaime pay?

$818.4

(b) What is the APR to the nearest half percent ?

8.5%

Step-by-step explanation:

(a) What finance charge will Jaime pay?

The cash cost is $7500. She decides to finance the project by paying 20% down

Amount paid for down payment = $7500 × 20/100

= $1500

Amount left = $7500 - $1500

= $6000

The balance paid in 36 monthly payments of $189.40.

Hence: $189.40 × 36 = $6818.4

The finance charge = $6818.4 - $6000

= $818.4

(b) What is the APR to the nearest half percent ?

Step 1

Finance charge/ Amount left × 100

= 818.4/6000 × 100

= 13.64

Using the APR table:

The APR to the nearest half percent = 8.5%


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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

A) y - 11 = -6(x - 0)

C) y + 1 = -6(x - 2)

Step-by-step explanation:

Plug in the corresponding x- and y- values in the table into the equations given in the answer choices.

I am going to use (1, 5) from the table.

A) y - 11 = -6(x - 0)[tex](5)-11=-6((1)-0)[/tex][tex]-6=-6(1)[/tex][tex]-6=-6[/tex]

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B) y - 1 = -6(x + 5)[tex](5)-1=-6((1)+5)[/tex][tex]4=-6(6)[/tex][tex]4 \not=-36[/tex]

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Answers

Answer:

see below

Step-by-step explanation:

Angles opposite each other when two lines cross. (see attached)

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Step-by-step explanation:

[tex]

\underline{\bf{Given\::}}

Given:

\underline{\bf{To\:find\::}}

Tofind:

\underline{\bf{Explanation\::}}

Explanation:

\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}

\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered}

−10

1

=

v

1

−30

1

v

1

=

−10

1

+

30

1

v

1

=

30

−3+1

v

1

=

30

−2

v

1

=

−15

1

⟶v=−15cm

\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}}

MAGNIFICATION:

\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered}

↦m=

Heightofobject(O)

Heightofimage(I)

=

Distanceofobject

Distanceofimage

=

u

v

↦m=

−15

−30

↦m=2cm

Thus;

The magnification will be 2 cm .

[/tex]

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