-22 - x = 5 + 6x + 9 Combine like terms (5 and 9)
-22 - x = 14 + 6x Subtract 6x from both sides
-22 - 7x = 14 Add 22 to both sides
-7x = 36 Divide both sides by -7
x = 5 $\frac{1}{7}$

6 0

3 years ago

(5+6b^3)^2 expand and combine like terms

Drupady [299]

Answer:

$( 5 + 6b^3)^2 = 36b^6 + 60b^3 + 25$

Step-by-step explanation:

$(a + b)^2 = a^2 + b^2 + 2ab$

So,

$(5 + 6b^3)^2 = (5)^2 + (6b^3)^2 + 2 (5)(6b^3)$

$= 25 + 36b^6 + 60b^3$

7 0

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

7 0

3 years ago