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vesna_86 [32]
3 years ago
6

(15–7xy^2)–(y^3–y^2x–15)

Mathematics
2 answers:
Vitek1552 [10]3 years ago
6 0
My answer is: -6xy2 - y3<span> + 30
Hope this helps bye</span>
blsea [12.9K]3 years ago
5 0
15 - 7xy^2 - (y^3 - y^2x - 15)

Distribute negative sign
15 - 7xy^2 + -1(y^3 - y2x - 15)
15 + -7xy^2 + -1y^3 + -1(-xy^2) + (-1)(-15)
15 + -7xy^2 + -y^3 + x^2 + 15

Combine like terms
15 + - 7xy^2 + -y^3 +xy^2 + 15
( -7xy^2 + xy^2) + (-y^3) + (15 + 15)
-6xy^2 + -y^3+ 30

= -6xy^2 + -y^3+ 30
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AB = 6 cm, AC = 12 cm, CD = ?

In triangle ABC, ∠CBA = 90°, therefore in triangle BCD ∠CBD = 90° also.

Since ∠BDC = 55°, ∠CBD = 90°, and there are 180 degrees in a triangle, we know ∠DCB = 180 - 55 - 90 = 35°

In order to find ∠BCA, use the law of sines:
 
sin(∠BCA)/BA = sin(∠CBA)/CA
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We know the sum of all angles in a triangle must be 180°, so we choose the value 30° for ∠BCA

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Since triangle DCA has 180°, we know ∠CAD = 180 - ∠DCA - ∠ADC = 180 - 65 - 55 = 60°

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