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PolarNik [594]
4 years ago
5

If PS = 2x + 18 and QR = 5x - 9, solve for x.

Mathematics
1 answer:
Alexxx [7]4 years ago
4 0

Answer:

2x+18+5x-9

2x+5x18-9

<u>7</u><u>x</u>+<u>9</u>

7. 7

x= 1.4

hope it helps

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madreJ [45]
Line DE is 4 squares and Line BC is 8 squares.

The length of line DE is 1/2 of line BC

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3 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3 5 , KE = 8, LM KN = 3 5 Find: KM, LM, KN, Area of KLMN
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
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3 years ago
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dem82 [27]

Answer:

the formula to find slope is m=y1-y2/x1-x2

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5 0
3 years ago
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bagirrra123 [75]

Answer:

x = 116° , y = 20°

Step-by-step explanation:

Since, measure of angle subtended on the center of the circle is twice the measure of angle subtended on the circumference of the circle.

\therefore \: x = 2 \times 58 \degree \\ \huge \red{ \boxed{  \therefore \: x =116 \degree}} \\   \\

Now, by angle sum property of a quadrilateral.

y + (360 - 116) \degree + 38 \degree + 58 \degree = 360 \degree \\ y + 244 \degree + 96 \degree = 360 \degree \\ y + 340 \degree= 360 \degree \\ y =  360 \degree - 340 \degree \\ \huge \purple{ \boxed{ y = 20 \degree}}

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4 years ago
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Solve for x and y simultaneously <br><br>Pls help me​
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Are the roots equal or unequal

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