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Sholpan [36]
3 years ago
12

What are the measurements for c, d & e

Mathematics
1 answer:
wel3 years ago
7 0
C = 180 - 53
c = 127

d = 540 - (127 + 114 + 92 + 119 )
d = 540 - 452
d = 88

e = 180 - 88
e = 92
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Then cross multiply the numbers and set up as a equation like this

(5/8) × 1 = (2/5) × r

(5/8) = (2/5)r

Multiply both sides of the equation by (5/2) to get rid of the fraction and solve for r

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25/16 = r

Reduce down to 5/4 or change to a mixed number

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4 years ago
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Step-by-step explanation:

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6 0
3 years ago
Factor x3y6 + 27z42. (xy3 + 3z39)(x2y9 – 3xy3z39 + 9z1,521) (xy3 + 3z39)(x2y6 – 3xy3z39 + 9z78) (x3y6 + 27z42)(x6y12 – 27x3y6z42
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Answer:

x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})

Step-by-step explanation:

Given

x^3y^6 + 27z^{42

Required

Factor

Express both terms as a cube

x^3y^6 + 27z^{42} = (xy^2)^3 + (3z^{14})^3

Let

a =(xy^2)

b = (3z^{14})

So:

x^3y^6 + 27z^{42} = a^3 + b^3

Using sum of cubes:

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

So:

x^3y^6 + 27z^{42} = (a + b)(a^2 - ab + b^2)

Substitute in, values of a and b

x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})((xy^2)^2 - (xy^2)*(3z^{14}) + (3z^{14})^2)

x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})

5 0
3 years ago
Read 2 more answers
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