Look at choice (B). The distance from (8,4) to the line with the equation x = 5 is |5 − 8| = |−3| =
3 . The distance from (8,4) to the line with the equation y = 7 is |7 − 4| = |3| = 3. The distances
from (3,7) to the lines with the equations x = 5 and y = 7 are equal; (B) is correct.35 . C
You should determine the range of each part of the piecewise function. Begin with x < 0. If
this is true, then 5 x < 0 and 5 x + 3 < 3. As x becomes more and more negative, 5 x + 3 will also
become more and more negative, so the range for this part of the function is all real values
less than 3.
Now consider x ≥ 0. Because the square of a number is nonnegative, x^2 ≥ 0 and x^2 + 1 ≥ 1.
Dividing both sides by x^2 + 1 results in . The range of is (0, 1], since x = 0
results in 1 and as x increases, the fraction will decrease but always remain positive.
Therefore, the range of the original function is , or (1, 3]. Combining these
two ranges gives the range of the full function as all real values less than or equal to 3, or y ≤
3.36 . D
Since you want the value of y, solve the equation x − y = 1 for x in terms of y. Then substitute
for x the expression containing y that equals x in the equation x^2 − 7y = 8. Adding y to both
sides of the equation x − y = 1, you have x = y + 1. Now substitute y + 1 for x in the equation x^2
− 7 y = 8. You have (y + 1)^2 − 7y = 8. Solve this equation for y.
The general solution to the quadratic equation ax^2 + bx + c = 0 isHere, you have the equation y^2 − 5y − 7 = 0 in the variable y. This equation is also a quadratic
equation where the right side is 0. You have the quadratic equation ay^2 + by + c = 0 where a =
1 , b = −5, and c = 7. The solution to this equation is