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Andreas93 [3]
3 years ago
5

The slope of a line parallel to the line who is the equation 2x + y = 6

Mathematics
1 answer:
Tanya [424]3 years ago
3 0

Answer:

-2

Step-by-step explanation:

Note that slope-intercept form looks like: y = mx + b

In this case, isolate the y. Note the equal sign, what you do to one side, you do to the other. Subtract 2x from both sides:

2x (-2x) + y = 6 (-2x)

y = -2x + 6

Note in the slope-intercept form: y = mx + b

y = y

m = slope

x = x

b = y-intercept

In this case, your slope is -2. All lines parallel to the given equation will also have a slope of-2.

~

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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:

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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.

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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

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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

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

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

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