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amid [387]
3 years ago
8

Colin buys a car for £45500. It decreases in price 6% per year. How much will it be worth in 3 years?

Mathematics
1 answer:
just olya [345]3 years ago
5 0

Answer:

The answer is 33,710 euros.

Step-by-step explanation:

Start by multiplying 45,550 by 6%. This will give 2,730. over the next three years the price decreases, so multiply 2,730 by 3, which is 8,190. Finally, subtract 45,550 by 8,190 to get 33,710.

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If 3 + root8 by 3-root8 + 3-root8 by 3+root8 = a+broot2 find a and b
s2008m [1.1K]

If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables or numbers.

\frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }\\ \\\frac{9+6\sqrt{8}+8+9-6\sqrt{8}+8  }{(3-\sqrt{8} )(3+\sqrt{8} )} \\\\\frac{34}{9-8} =34+0\sqrt{2}

If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

Find out more on equation at: brainly.com/question/2972832

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8 0
2 years ago
Solve for x, y, and z
valentina_108 [34]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Help please i need to pass
stellarik [79]

Answer:

hi! i dont have the answer because i suck at school lol XD but, i actually have the same question! do you know the answer yet?:] sorry to bother you if you dont.

Step-by-step explanation:

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3 years ago
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Square A has a side length of (2x - 7) and Square B has a side length (-4x + 18). How much bigger is the perimeter of Square B t
stich3 [128]

Answer: 100 - 24x

Step-by-step explanation:

The formula for calculating the perimeter of a square is given by :

P = 4l , where l is the length of the side.

Perimeter of square A will be

P = 4 ( 2x - 7 )

P = 8x - 28

Perimeter of square B will be ;

P = 4 ( -4x + 18 )

P = - 16x + 72

Perimeter of square B - Perimeter of square A implies

-16x + 72 - ( 8x - 28 )

-16x + 72 - 8x + 28

collecting the like terms

-16x - 8x + 72 + 28

-24x + 100

⇒ 100 - 24x

3 0
3 years ago
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