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solong [7]
3 years ago
13

Solve for p.

Mathematics
2 answers:
Marat540 [252]3 years ago
8 0
Hi there

3(p + q) = q
Solve for P

Use the distributive property
(3)(p) + (3)(q) = q
3p + 3q = q
3p = q-3q
3p = -2q
p = -2q/3

The answer is D

If you have any further questions please let me know :)
Andrej [43]3 years ago
8 0
3(p+q) = p

First we want to break apart the brackets. The 3 can be distributed as a coefficient to p and q .

3p + 3q = 1p

Next we want to isolate the variable p to a single side of the equation

3p + 3q - 3p = 1p - 3p

Simplifying the equation, we get

3q = -2p

Isolating for p, we get: 3q/-2 = p

This answer corresponds to answer D.
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In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
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