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Whitepunk [10]
3 years ago
10

4C. Quintin is using the three different shaped

Mathematics
1 answer:
Sauron [17]3 years ago
3 0

The smallest number of tiles Quintin will need in order to tile  his floor is 20

The given parameters;

  • number of different shapes of tiles available = 3
  • number of each shape = 5
  • area of each square shape tiles, A = 2000 cm²
  • length of the floor, L = 10 m = 1000 cm
  • width of the floor, W = 6 m = 600 cm

To find:

  • the smallest number of tiles Quintin will need in order to tile his floor

Among the three different shapes available, total area of one is calculated as;

A_{one \ square \ type} = 5 \times 2000 \ cm^2 = 10,000 \ cm^2

Area of the floor is calculated as;

A_{floor} = 1000 \ cm \times 600 \ cm = 600,000 \ cm^2

The maximum number tiles needed (this will be possible if only one shape type is used)

maximum \ number= \frac{Area \ of \ floor}{total \ area \ of \ one \ shape \ type} \\\\maximum \ number= \frac{600,000 \ cm^2}{10,000 \ cm^2} \\\\maximum \ number=  60

When all the three different shape types are used we can get the smallest number of tiles needed.

The minimum or smallest number of tiles needed (this will be possible if all the 3 different shapes are used)

3 \times \ smallest \ number  = 60\\\\smallest \ number = \frac{60}{3} \\\\smallest \ number = 20

Thus, the smallest number of tiles Quintin will need in order to tile  his floor is 20

Learn more here: brainly.com/question/13877427

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Please show work if you do, ( and get it right) I will make you brainliest!!!
DerKrebs [107]

Answer:

The percentage increase in bird population in the second years is 4%  .

Step-by-step explanation:

Given as :

During two years, the population of birds of an island became 7 times more than before.

The the percentage increase in first year = = r_1 =40%

Let The percentage increase in second year =  r_2 = r%

So, According to question

The initial population of bird = x

The increase population of bird =7 times before = 7 x

Now,

The increase population of bird = initial population of bird × (1+ \dfrac{r_1}{100}) ×  (1+ \dfrac{r_2}{100})

Or, 7 x = x × (1+ \dfrac{40}{100}) ×  (1+ \dfrac{r}{100})

Or, \dfrac{7 x}{x} = 1.4 ×  (1+ \dfrac{r}{100})

Or, 7 = 1.4 ×  (1+ \dfrac{r}{100})

Or, \dfrac{7 }{1.4} = (1+ \dfrac{r}{100})

Or, 5 = (1+ \dfrac{r}{100})

Or, (1+ \dfrac{r}{100}) = 5

Or, \dfrac{r}{100} = 5 - 1

Or,  \dfrac{r}{100} = 4

∴ r = 4 × 100

I.e r = 400

so, The percentage increase = r% = 4

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Which formula can be used to find the nth term of the geometric sequence below?
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Answer:

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Step-by-step explanation:

The first term a1 = 1/6 and the common  ratio r =  1 / 1/6 = 6, also 36/6 = 6 which confirms this.

So the nth term = a1 r^(n - 1)

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4 years ago
Match the cards to show the equivalencies.
Nadya [2.5K]

Answer:

I dont know

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. Anne is pushing a wheelbarrow filled with mulch to place in her garden. She is pushing the wheelbarrow with a force of 70 N at
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<u>Answer:</u>

When Anne pushes the wheelbarrow 25 meters the work done is 1120 joules

<u>Solution:</u>

We know \text { Work }(\mathrm{W})=\text { Force }(\mathrm{F})\times\text { Distance (D) }

Anne is pushing the wheelbarrow with a force of 70 N at an angle of 50^{\circ} with the horizontal

So the force of Anne will be F=70 \cos 50^{\circ}=(70 \times 0.64)=44.8 \mathrm{N}

She pushes the wheelbarrow 25 meter, So D = 25 meters

So, the work,F\times D=(44.8 \times 25)=1120 \text { Joules }

Anne has done total 1120 joules work.

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