Solve and give your answer in interval notation 6w 18 lesst

Solve and give your answer in interval notation: |6w + 18| lessthanorequalto 6 [-4, 4] [-4, -2] [2, 4] [-2, 2] A line passes through the point (8, 3) and has a slope of 5/4 write an equation in slope-intercept form for this line. Y = 5/4x - 3/8 y = 5/4x - 5 x = 4/5y + 3/8 y = 5/4x - 7 Consider the line y = -5x+9 Find the equation of the line that is parallel to this line and passes through the point (3, 6) y = -5x + 21 y = 1/5x + 27/5 y = -5x + 6 y = -1/5x + 9/5 The credit remaining on a phone card (in dollars) is a linear function of the total calling time made card (in minutes). The remaining credit after 32 minutes of calls is $25.20 and the remaining credit 66 minutes of calls is $20.10. What is the remaining credit after 70 minutes of calls? $10.20 $18.70 $19.50 $26

Solution

4. We have |6w + 18 | 6 so that – 6 ( 6w + 18) 6 or, (-6 – 18 ) 6w 6 – 18 or, - 24 6w -12.Now , on dividing by 6, we get -4 w -2. Thus w [ -4, -2]. The option b) is the correct answer.

5. The slope-intercept form of the equation of a line is y = mx + c where m is the slope and c is the y –intercept. Here, the slope is 5/4 so that the equation of the line changes to y = (5/4)x + c. Further, since the line passes through the point (8, 3), on substituting x = 8 and y = 3 in the equation of the line, we have 3 = (5/4)* 8 + c or, 3 = 10 + c so that c = 3 – 10 = -7. Thus, the equation of the required line is     y = (5/4) x – 7. The option d) is the correct answer.

6. We know that two parallel lines have the same slope. Therefore, the slope of the required line is -5. Let the equation of the required line be y = -5x + c where c is its y – intercept. Since this line passes through the point (3, 6), on substituting x = 3 and y = 6 in the equation of the line, we have 6 = -5*3 + c or, c = 6 + 15 = 21. Thus, the equation of the required line is y = -5 x +21. The option a) is the correct answer.

7. Let the equation of the remaining credit on the phone card be y = ax + b, where y is the remaining credit , x is the minutes of calls made with the phone card and a and b are arbitrary real numbers. The remaining credit on the phone card after 32 minutes of calls is $ 25.20 so that 25.20 = 32a + b…(1). Also, since the remaining credit on the phone card after 66 minutes of calls is $ 20.10 , we have 20.10 = 66a + b…(2) On subtracting the 1^st equation from the 2^nd equation, we get 66a + b – 32a – b = 20.10 – 25.20 or, 34a = -5.10 so that a = -5.10/34 =- 0.15. Now, on substituting a = -0.15 in the 1^st equation, we get 25.20 = 32*( -0.15) + b or, 25.20 = - 4.80 + b or, b = 25.20 + 4.80 = 30. Thus, the required equation is y = - 0.15x + 30. On substituting x = 70, we have y = -0.15 *70 + 30 = - 10.50 + 30 = 19.50. Thus, the remaining credit on the phone card after 70 minutes of calls is $ 19.50. The option c) is the correct answer.