The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The size of angle C is 59o. Find the length of side AC.
Let x be the length of side AC. Use the cosine law 122 = 82 + x2 - 2*8*x*cos(59o) Solve the quadratic equation for x: x = 14.0 and x = -5.7 x cannot be negative and therefore the solution is x = 14.0 (rounded to one decimal place).
ABCD is a parallelogram such that AB is parallel to DC and DA parallel to CB. The length of side AB is 20 cm. E is a point between A and B such that the length of AE is 3 cm. F is a point between points D and C. Find the length of DF such that the segment EF divide the parallelogram in two regions with equal areas.
Let A1 be the area of the trapezoid AEFD. Hence A1 = (1/2) h (AE + DF) = (1/2) h (3 + DF) , h is the height of the parallelogram. Now let A2 be the area of the trapezoid EBCF. Hence A2 = (1/2) h (EB + FC) We also have EB = 20 - AE = 17 , FC = 20 - DF We now substitute EB and FC in A2 = (1/2) h (EB + FC) A2 = (1/2) h (17 + 20 - DF) = (1/2) h (37 - DF) For EF to divide the parallelogram into two regions of equal ares, we need to have area A1 and area A2 equal (1/2) h (3 + DF) = (1/2) h (37 - DF) Multiply both sides by 2 and divide thm by h to simplify to 3 + DF = 37 - DF Solve for DF 2DF = 37 - 3 2DF = 34 DF = 17 cm
solve the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13
Given the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13 Multiply factors. -15x - 10 - x + 3 = -16x - 20 +13 -16x - 7 = -16x - 7 Add 16x + 7 to both sides and write the equation as follows 0 = 0 The above statement is true for all values of x and therefore all real numbers are solutions to the given equation.