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The equation (y-2)=(1)/(5)(x+5) is graphed in the xy-plane. Which of the statements below is true of its graph? Choose 1 answer: A The graph has a slope of -5 and a y-intercept of 5 . (B) The graph has a slope of -5 and passes through the point (2,-5). (C) The graph has a slope of (1)/(5) and passes through the point (-5,2). D The graph has a slope of (1)/(5) and a y-intercept of 5 .

The equation\newline(y−2)=15(x+5) (y-2)=\frac{1}{5}(x+5) is graphed in the xy x y -plane. Which of the statements below is true of its graph?\newlineChoose 11 answer:\newline(A) The graph has a slope of −5-5 and a y y -intercept of 55 .\newline(B) The graph has a slope of −5-5 and passes through the point (2,−5) (2,-5) .\newline(C) The graph has a slope of 15 \frac{1}{5} and passes through the point (−5,2) (-5,2) .\newlineD The graph has a slope of 15 \frac{1}{5} and a y y -intercept of 55 .

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Full solution

Q. The equation\newline(y−2)=15(x+5) (y-2)=\frac{1}{5}(x+5) is graphed in the xy x y -plane. Which of the statements below is true of its graph?\newlineChoose 11 answer:\newline(A) The graph has a slope of −5-5 and a y y -intercept of 55 .\newline(B) The graph has a slope of −5-5 and passes through the point (2,−5) (2,-5) .\newline(C) The graph has a slope of 15 \frac{1}{5} and passes through the point (−5,2) (-5,2) .\newlineD The graph has a slope of 15 \frac{1}{5} and a y y -intercept of 55 .
  1. Identify slope and intercepts: We need to identify the slope and intercepts of the given equation to determine which statement is true. The equation is given in a form that is close to slope-intercept form, which is y=mx+by = mx + b, where mm is the slope and bb is the yy-intercept.
  2. Rewrite equation in slope-intercept form: First, let's rewrite the equation in slope-intercept form by isolating yy on one side of the equation.y−2=(15)(x+5)y - 2 = \left(\frac{1}{5}\right)(x + 5)Add 22 to both sides to isolate yy.y=(15)(x+5)+2y = \left(\frac{1}{5}\right)(x + 5) + 2
  3. Distribute and combine terms: Now, let's distribute the (15)(\frac{1}{5}) across the terms inside the parentheses.\newliney=(15)x+(15)⋅5+2y = (\frac{1}{5})x + (\frac{1}{5})\cdot5 + 2\newliney=(15)x+1+2y = (\frac{1}{5})x + 1 + 2
  4. Simplify the equation: Combine the constant terms to find the y-intercept.\newliney=15x+3y = \frac{1}{5}x + 3\newlineThis equation tells us that the slope (mm) is 15\frac{1}{5} and the y-intercept (bb) is 33.
  5. Compare findings with answer choices: Now, let's compare our findings with the answer choices.\newlineA) Incorrect, because the slope is not −5-5 and the y-intercept is not 55.\newlineB) Incorrect, because the slope is not −5-5 and it does not pass through the point (2,−5)(2, -5).\newlineC) Correct, because the slope is 15\frac{1}{5} and the equation passes through the point (−5,2)(-5, 2).\newlineD) Incorrect, because the y-intercept is not 55, it is 33.
  6. Final Answer: True statement is: \newlineC) The graph has a slope of 15 \frac{1}{5} and passes through the point (−5,2) (-5,2) .

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