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oee [108]
3 years ago
9

Combine like terms: 3p2q2-3p2q3+4p2q3-3p2q2+pq PLEASE HELP!!! ASAP!!!

Mathematics
1 answer:
ch4aika [34]3 years ago
7 0

Answer:

p²q³ + pq and pq(pq² + 1)

Step-by-step explanation:

Given

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Required

Collect like terms

We start by rewriting the expression

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Collect like terms

3p²q² -3p²q² - 3p²q³ +4p²q³ + pq

Group like terms

(3p²q² -3p²q²) - (3p²q³ - 4p²q³ ) + pq

Perform arithmetic operations on like terms

(0) - (-p²q³) + pq

- (-p²q³) + pq

Open bracket

p²q³ + pq

The answer can be further simplified

Factorize p²q³ + pq

pq(pq² + 1)

Hence, 3p²q² - 3p²q³ +4p²q³ -3p²q² + pq is equivalent to p²q³ + pq and pq(pq² + 1)

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Answer:

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Area: T = 6444.961

Perimeter: p = 366

Semiperimeter: s = 183

Angle ∠ A = α = 60° = 1.047 rad

Angle ∠ B = β = 60° = 1.047 rad

Angle ∠ C = γ = 60° = 1.047 rad

Height: ha = 105.655

Height: hb = 105.655

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Median: ma = 105.655

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Inradius: r = 35.218

Circumradius: R = 70.437

Vertex coordinates: A[122; 0] B[0; 0] C[61; 105.655]

Centroid: CG[61; 35.218]

Coordinates of the circumscribed circle: U[61; 35.218]

Coordinates of the inscribed circle: I[61; 35.218]

Exterior (or external, outer) angles of the triangle:

∠ A' = α' = 120° = 1.047 rad

∠ B' = β' = 120° = 1.047 rad

∠ C' = γ' = 120° = 1.047 rad

Step-by-step explanation:

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