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MissTica
2 years ago
15

Two figures are similar with a scale factor of 5/2.

Mathematics
1 answer:
aev [14]2 years ago
6 0

Answer:

If we have two figures, F and F'

Such that if we start with F, and dilate it with a scale factor K, we get F'.

We will have:

All the measures of F', are K times the correspondent measures of F.

Then if F has  s₁, s₂, ..., sₙ sides, the sides of F' will be:

K*s₁, K*s₂, ..., K*sₙ

The ratio between correspondent sides will be equal to K

The ratio between perimeters will also be equal to K (because the perimeter is the sum of all the sides of each figure, so we can just take K as a common factor)

In the case of the area, because we usually multiply a measure by another, a factor K^2 will  appear, and the quotient between the areas is K^2

And finally, for the volumes, the ratio will be K^3

a) The ratio of corresponding lengths is K, in this case is 5/2

b) The ratio of the perimeters is K, in this case is 5/2

c) The ratio of the areas is K^2, in this case is (5/2)^2 = 25/4

d) The ratio of the areas is K^3, in this case is (5/2)^3 = 125/8

e) Two figures are similar if the figures have the same shape, then the corresponding angles are exactly the same, then the ratio of corresponding angle measures is 1.

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a bag contains 10 white gold balls and 6 striped golf balls. a golfer wants to add 112 golf balls to the bag. he wants the ratio
wolverine [178]

Answer:70 white golf balls and 42 striped golf balls

Step-by-step explanation:

First we find the ratio of white to striped balls

White golf balls = 10

Striped golf balls = 6

Ratio = 10/6 = 5/3

We are told a golfer wants to add extra 112 balls to the already 16 balls

And the ratio after adding 112 balls must stay the same

First we label the extra golf balls to be added x and y

x = white golf balls

y = striped golf balls

So since we know the 112 balls added is a combination of the extra white golf balls and striped golf balls, we create an equation for that, labelling it (1)

x + y = 112 (1)

And we are told that after putting these extra balls the ratio must remain the same, which is 5/3

which will be (10 white balls + x) divided by (6 striped ball + y) will be equals to 5/3

So we create another equation for this, labelling it (2)

(10+x)/(6+y) = 5/3 (2)

So we have two simultaneous equations

We pick (1)

x + y = 112

We either make x or y the subject of formula, I choose to make x the subject of formula, we label the equation (3)

take y to the other side, causing it to change to -y

x = 112 - y (3)

We then work with (2)

(10+x)/(6+y) = 5/3

We cross multiply

3(10+x) = 5(6+y)

We open the brackets

Making the equation simplified and labelling it (4)

30 +3x = 30 + 5y

Collect like terms

3x -5y = 30-30

3x -5y = 0 (4)

Remember from (3) we know that

x = 112 -y

So we put (3) in (4)

3(112 - y) - 5y = 0

Open bracket

336 -3y -5y =0

336 -8y = 0

Transfer -8y to the other side, changing to +8y

336 = 8y

Divide both sides by 8

336/8 = y

42 = y

y = 42

from (3) we know that x equals 112 - y

So we put y = 42 in (3)

x = 112 - y

x = 112 -42 = 70

x = 70

So therefore number of white golfs balls and striped golfs balls to be added to keep the same ratio is 70 and 42 respectively

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