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Licemer1 [7]
1 year ago
10

PROVING FIGURES SIMILAR please help! :)

Mathematics
2 answers:
Alja [10]1 year ago
7 0

a) The dimensions that give similar figures are:

8 in by 10 in and 16in by 20 in

b) If we resize the figures between the two above sizes, the figure will only increase its size and will not change in any other way.

<h3>how to identify similar figures?</h3>

Remember that two rectangles are similar if the quotient between the lengths and widths gives the same for both rectangles.

For the most common picture size, the dimensions are 4in by 6 in, the quotient between the dimensions gives:

6in/4in = 3/2.

Now, if we get the quotient for all the other options we will get:

5 in by 7 in:   Q = 7/5

8 in by 10 in:  Q = 10/8 = 5/4

9 in by 12 in: Q = 12/9 = 4/3

11 in by 14in: Q = 14/11

14in by 18in: Q = 18/14 = 9/7

16in by 20 in: Q = 20/16 = 10/8 = 5/4

Thus, we can conclude that the only pair of similar rectangles are 8 in by 10 in and 16in by 20 in

b) What can we conclude from that?

It means that if we change the size from  8 in by 10 in to 16in by 20 in, the figure will not change, it will only increase its size (for the other sets of dimensions we would see that the figure gets distorted, as in these cases the rectangles are not similar).

If you want to learn more about similar figures:

brainly.com/question/14285697

#SPJ1

borishaifa [10]1 year ago
4 0

Answer:

8x10 and 16x20

Step-by-step explanation:

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so
given
we want 5% left
so A=5% of P or 5/100 of P or 1/20 of P or 0.05P

h=4.6

so
0.05P=P(\frac{1}{2})^\frac{t}{4.6}
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take ln of both sides
ln(0.05)=ln(\frac{1}{2})^\frac{t}{4.6}
ln(0.05)=(\frac{t}{4.6})ln(\frac{1}{2})
ln(0.05)=(\frac{t}{4.6})ln(\frac{1}{2})
divide both sides by ln(\frac{1}{2})
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Read 2 more answers
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