Ageometry class is asked to find the equation of a line that is parallel to y – 3 = −(x + 1) and passes through (4, 2). trish states that the parallel line is y – 2 = –1(x – 4). demetri states that the parallel line is y = –x + 6. are the students correct? explain. trish is the only student who is correct; the slope should be –1, and the line passes through (4, 2). demetri is the only student who is correct; the slope should be –1, and the y-intercept is 6. both students are correct; the slope should be –1, passing through (4, 2) with a y-intercept of 6. neither student is correct; the slope of the parallel line should be 1.

Both students are correct; the slope should be –1, passing through (4, 2) with a y-intercept of 6

Step-by-step explanation:

Two equations are parallel when they have the same slope. Given the function:

y – 3 = −(x + 1)

It has the form (called point-slope form): y - y0 = m*(x -x0)

where (x0,y0) is a point that correspond to the line and m is the slope,  so the its slope is -1.

Then the equation:

y - 2 = -(x - 4)

is parallel to the original and passes through (4, 2). So, Trish is correct.

Another way to express the equation of a line is the slope-intercept form:

y = m*x + b

where b is the y-intercept of the equation. The slope has to be -1 and the equation has to passes through (4, 2),  then:

2 = (-1)*4 + b

2 + 4 = b

b = 6

Therefore, the equation y = -x + 6 satisfy the requirements, and Demetri was also correct.

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.