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riadik2000 [5.3K]
3 years ago
6

5pn-8p=7n-7. Solve for n.

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

       -8p + 7

n = --------------

        8 - 5p

Step-by-step explanation:

Let's group the n terms in  5pn-8p=7n-7  .

Subtract 5pn from both sides.  This results in -8p = 7n - 7 - 5pn

Now add 7 to both sides.  This will group the n terms on the right side:

-8p + 7 = 7n - 5pn

Factor n out of 7n - 5pn:

-8p + 7 = n(7 - 5p)

Finally, divide both sides by (7 - 5n) to isolate n:

       -8p + 7

n = --------------

        8 - 5p

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one morning the temperature was 3⁰f below zero.at noon the temperature has risen 12⁰f and then dropped 6⁰f by 6:00 that evening.
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Answer:

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4 years ago
ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl
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Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

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Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x}  \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

\frac{AQ}{QP}=\frac{AB}{CP}  \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19

\frac{BQ}{CQ}=\frac{AB}{CP}  \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

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PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

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