Answers for the 2 boxes please
1 answer:
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MO = 12 and PR = 3
Solution:
Given .
Perimeter of ΔMNO = 48
Perimeter of ΔPQR = 12
MO = 12x and PR = x + 2
<em>If two triangles are similar, then the ratio of corresponding sides is equal to the ratio of perimeter of the triangles.</em>
Do cross multiplication.
Subtract 48x from both sides.
Divide by 96 on both sides, we get
⇒ 1 = x
⇒ x = 1
Substitute x = 1 in MO an PR.
MO = 12(1) = 12
PR = 1 + 2 = 3
Therefore MO = 12 and PR = 3.
<h3>Given</h3>
72 blue, 42 red beads
beads are used to make identical necklaces
<h3>Find</h3>(a) the greatest number of necklaces that can be made
(b) the number of each color bead in each necklace
<h3>Solution</h3>You can write and factor the equation
... necklaces = 72 blue + 42 red
... necklaces = 6(12 blue + 7 red)
where 6 is the greatest common factor (GCF) of 72 and 42.
(a) You can make up to 6 identical necklaces. 6 is the largest common factor of 72 and 42. If you were to try to make more, they could not be identical.
(b) Each necklace can consist of 12 blue and 7 red beads. These are the numbers obtained when the total bead count is divided into 6 equal groups.
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There are several ways you can find the GCF of two numbers. For small numbers, it is generally feasible to use your knowledge of multiplication tables and factors to choose the largest common factor of two numbers. You can also use Euclid's algorithm, which is to repeatedly compute
... (largest number) modulo (smallest number)
until the result is zero. The final "smallest number" is the GCF.
Here, that looks like
... 72 mod 42 = 30
... 42 mod 30 = 12
... 30 mod 12 = 6
... 12 mod 6 = 0 . . . . . . . so 6 is the GCF
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Of course, you know that
... 72 = 2³×3²
... 42 = 2×3×7
so, the largest set of common factors is 2×3 = 6.
___
Your graphing calculator may have a function for computing the greatest common divisor (GCD), too. The TI-84 does, for example.