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Anuta_ua [19.1K]
3 years ago
14

You have decided to purchase a car for $25,625. The credit union requires a 10% down payment and will finance the balance

Mathematics
2 answers:
katrin2010 [14]3 years ago
5 0

Answer

c. $27,722.01

I just took the test :)

swat323 years ago
4 0

Answer:

Total purchase = $28498.685

In cent = 2849868.5 cent.

Step-by-step explanation:

Given : You have decided to purchase a car for 175.13.

To find : What is the total purchase price of the car including tax, license, and title?

Solution :

Price of car = $25,625

Sales tax = 7.5% = 0.075

The total purchase price of the car including tax is

T=25625\times 0.075=\$1921.875T=25625×0.075=$1921.875

Title charges are $175.13.

The credit union requires a 10% down payment and will finance the balance with a 9% annual interest loan for 36 months.

10% down payment on price of car

25625\times 0.10=2562.525625×0.10=2562.5

25625-2562.5=23062.525625−2562.5=23062.5

The amount will be $23062.5

Now, we find the monthly payment,

Monthly payment, M=\frac{\text{Amount}}{\text{Discount factor}}M=

Discount factor

Amount

​

Discount factor D=\frac{1-(1+i)^{-n}}{i}D=

i

1−(1+i)

−n

​

Rate = 9%=0.09

i=\frac{0.09}{12}=0.0075i=

12

0.09

​

=0.0075

Time n=36 month

Now, put all the values we get,

D=\frac{1-(1+i)^{-n}}{i}D=

i

1−(1+i)

−n

​

D=\frac{1-(1+0.0075)^{-36}}{0.0075}D=

0.0075

1−(1+0.0075)

−36

​

D=\frac{1-(1.0075)^{-36}}{0.0075}D=

0.0075

1−(1.0075)

−36

​

D=\frac{0.2358}{0.0075}D=

0.0075

0.2358

​

D=31.44D=31.44

Monthly payment, M=\frac{\text{Amount}}{\text{Discount factor}}M=

Discount factor

Amount

​

M=\frac{23062.5}{31.44}M=

31.44

23062.5

​

M=733.38M=733.38

Monthly payment is $733.38

For 36 months the payment is 733.38\times 36=26401.68733.38×36=26401.68

Licence = $26401.68

The total purchase price of the car including tax, license, and title

Total purchase = 1921.875 +1921.875+175.13 +$26401.68

Total purchase = $28498.685

1 dollar = 100 cent

$28498.685 = 2849868.5 cent.

Step-by-step explanation:

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3 years ago
point b on the ground is 5 cm from point E at the entrance to Ollie's house. He is 1.8 m tall and is standing at Point D, below
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Point B on the ground is 5 cm from point E at the entrance to Ollie's house.

Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

The complete question is as follows:

Ollie has installed security lights on the side of his house that is activated by a  sensor. The sensor is located at point C directly above point D. The area covered by the sensor is shown by the shaded region enclosed by triangle ABC. The distance from A to B is 4.5 m, and the distance from B to C is 6m. Angle ACB is 15°.

The objective of this information is:

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The diagrammatic representation of the information given is shown in the image attached below.

Using  cosine rule to determine angle CAB, we have:

\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}= \dfrac{CA}{Sin \hat {ABC}}}

Here:

\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}}

\mathbf{\dfrac{4.5}{Sin \hat {15^0}} = \dfrac{6}{Sin \hat {CAB}}}

\mathbf{Sin \hat {CAB} = \dfrac{Sin 15 \times 6}{4.5}}

\mathbf{Sin \hat {CAB} = \dfrac{0.2588 \times 6}{4.5}}

\mathbf{Sin \hat {CAB} = 0.3451}

∠CAB = Sin⁻¹ (0.3451)

∠CAB = 20.19⁰

From the diagram attached;

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Then, we can say:

∠CBD = ∠GBF

∠GBF = (CAB + ACB)      

(because the exterior angles of a Δ is the sum of the two interior angles.

∠GBF = 15° + 20.19°

∠GBF = 35.19°

Using the trigonometric function for the tangent of an angle.

\mathbf{Tan \theta = \dfrac{GF}{BF}}

\mathbf{Tan \ 35.19  = \dfrac{1.8 \ m }{BF}}

\mathbf{BF  = \dfrac{1.8 \ m }{Tan \ 35.19}}

\mathbf{BF  = \dfrac{1.8 \ m }{0.7052}}

BF = 2.55 m

Finally, the distance of Ollie║FE║ from the entrance of his bouse is:

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8 0
3 years ago
Factorize :3y²-54y+343​
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−

b

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b

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4

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c

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b

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a(x−

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b

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3

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(

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54

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3

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2

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(

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​

)(y−

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​

)

3 Simplify.

3

(

y

−

54

+

20

3

ı

6

)

(

y

−

54

−

20

3

ı

6

)

3(y−

6

54+20

3

​



​

)(y−

6

54−20

3

​



​

)

4 Factor out the common term

2

2.

3

(

y

−

2

(

27

+

10

3

ı

)

6

)

(

y

−

54

−

20

3

ı

6

)

3(y−

6

2(27+10

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​

)

​

)(y−

6

54−20

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​



​

)

5 Simplify

2

(

27

+

10

3

ı

)

6

6

2(27+10

3

​

)

​

to

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+

10

3

ı

3

3

27+10

3

​



​

.

3

(

y

−

27

+

10

3

ı

3

)

(

y

−

54

−

20

3

ı

6

)

3(y−

3

27+10

3

​



​

)(y−

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

​

)

6 Factor out the common term

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3

(

y

−

27

+

10

3

ı

3

)

(

y

−

2

(

27

−

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3

ı

)

6

)

3(y−

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3

​



​

)(y−

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​

)

​

)

7 Simplify

2

(

27

−

10

3

ı

)

6

6

2(27−10

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​

)

​

to

27

−

10

3

ı

3

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

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.

3

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y

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+

10

3

ı

3

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(

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27

−

10

3

ı

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