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Keith_Richards [23]
3 years ago
5

Mr Blossom purchased a 2006 used Sedan. The cost of the car after a 5% rebate was $11,566. What was tge cost of the car before t

he rebates?
Mathematics
1 answer:
Naya [18.7K]3 years ago
4 0
Let's say the original price was "x", so that'd be the 100%, but he got a rebate of 5% off the regular price, and it ended up at 11566, so, 11566 is 100% - 5% or 95%.

now, if we know that 11566 is the 95% from the original price, what is "x"?

\bf \begin{array}{ccll}
a&b\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
x&100\\
11566&95
\end{array}\implies \cfrac{x}{11566}=\cfrac{100}{95}\implies x=\cfrac{11566\cdot 100}{95}
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Item 7
Mariulka [41]

Answer:

A = 74.7^\circ

B = 42.5^\circ

C = 62.8^\circ

Step-by-step explanation:

Given

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

For AB

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

d = \sqrt{(-1 - 2)^2 + (2 - 8)^2

d = \sqrt{(-3)^2 + (-6)^2

d = \sqrt{45

So:

AB = \sqrt{45

For BC

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

BC = \sqrt{(2 - 4)^2 + (8 - 1)^2

BC = \sqrt{(-2)^2 + (7)^2

BC = \sqrt{53

For AC

A = (-1,2) \to (x_1,y_1)

C = (4,1) \to (x_3,y_3)

AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2

AC = \sqrt{(-5)^2 + (1)^2

AC = \sqrt{26

So, we have:

AB = \sqrt{45

BC = \sqrt{53

AC = \sqrt{26

By representation

AB \to c

BC \to a

AC \to b

So, we have:

a = \sqrt{53

b = \sqrt{26

c = \sqrt{45

By cosine laws, the angles are calculated using:

a^2 = b^2 + c^2 -2bc \cos A

b^2 = a^2 + c^2 -2ac \cos B

c^2 = a^2 + b^2 -2ab\ cos C

a^2 = b^2 + c^2 -2bc \cos A

(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A

53 = 26 +45 - 2 * 34.21 * \cos A

53 = 26 +45 - 68.42 * \cos A

Collect like terms

53 - 26 -45 = - 68.42 * \cos A

-18 = - 68.42 * \cos A

Solve for \cos A

\cos A =\frac{-18}{-68.42}

\cos A =0.2631

Take arc cos of both sides

A =\cos^{-1}(0.2631)

A = 74.7^\circ

b^2 = a^2 + c^2 -2ac \cos B

(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B

26 = 53 +45 -97.67 * \cos B

Collect like terms

26 - 53 -45= -97.67 * \cos B

-72= -97.67 * \cos B

Solve for \cos B

\cos B = \frac{-72}{-97.67}

\cos B = 0.7372

Take arc cos of both sides

B = \cos^{-1}(0.7372)

B = 42.5^\circ

For the third angle, we use:

A + B + C = 180 --- angles in a triangle

Make C the subject

C = 180 - A -B

C = 180 - 74.7 -42.5

C = 62.8^\circ

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