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Anvisha [2.4K]
3 years ago
12

At the start of the week a bookshop had science and art books in the ratio 2:5. By the end of the week, 20% of each type of book

s were sold and 2240 books of both types were unsold. How many books of each type were there at the start of the week?
Mathematics
1 answer:
Anastaziya [24]3 years ago
5 0
Let there be 2x science and 5x art books 
<span>science books sold = 2x × 0.2 = 0.4x </span>
<span>science books unsold = 2x – 0.4x = 1.6x </span>

<span>art books sold = 5x × 0.2 = x </span>
<span>art books unsold = 5x – x = 4x </span>

<span>total books unsold = 1.6x + 4x = 5.6x </span>
<span>5.6x = 2240 </span>
<span>x = 400 </span>

<span>2x science = 800 </span>
<span>and 5x art books = 2000 </span>
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The table shows ordered pairs of the function y = 16 + 0.5x .
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The missing values represented by x and y are 8 and 20, that is

(x, y) = (8, 20)

The function y = 16 + 0.5x is a linear equation that can be solved graphically. This means the values of both variables x and y can be found on different points along the straight-line graph.

The ordered pairs simply mean for every value of x, there is a corresponding value of y.

The 2-column table has values for x and y which all satisfy the equation y = 16 + 0.5x. Taking the first row, for example, the pair is given as (-4, 14).

This means when x equals negative 4, y equals 14.

Where y = 16 + 0.5x

y = 16 + 0.5(-4)

y = 16 + (-2)

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y = 14

Therefore the first pair, just like the other four pairs all satisfy the equation.

Hence, looking at the options given, we can determine which satisfies the equation

(option 1) When x = 0

y = 16 + 0.5(0)

y = 16 + 0

y = 16

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(option 2) When x = 5

y = 16 + 0.5(5)

y = 16 + 2.5

y = 18.5

(5, 18.5)

(option 3) When x = 8

y = 16 + 0.5(8)

y = 16 + 4

y = 20

(8, 20)

From our calculations, the third option (8, 20) is the correct ordered pair that would fill in the missing values x and y.

To learn more about the straight line visit:

brainly.com/question/1852598

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Step-by-step explanation:

If two exponents have the same base and are multiplied together,  you will keep the base unchange and add the exponents.  

So looking at the equation, the base is 2 for both and they are raised to the powers  g and h, meaning that if  g and h are added together they have  to equal -3.

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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:

  • To find angle CAB and;
  • Find the distance Ollie is from the entrance to his house when he first activates the sensor.

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;

  • assuming we have an imaginary position at the base of Ollie Standing point called point F when Ollie first activates the sensor;          

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:

= 5 - 2.55 m

= 2.45 m

Therefore, we can conclude that Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

Learn more about exterior angles here:

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