Answer:
The surface area of a cylinder is the sum of the areas of the two bases and the area of the lateral surface. If the height of the cylinder is doubled, the area of the lateral surface is doubled, but the areas of the bases remain the same.
Step-by-step explanation:
Answer:
( - x, - y )
Step-by-step explanation:
The starting is always Quadrant I.
270 degrees clockwise from Quadrant I is Quadrant III.
In Quadrant III, the points will be in the form ( - x, - y ).
Answer:
The distance between the points is approximately 6.4
Step-by-step explanation:
The given coordinates of the points are;
(2, -2), and (6, 3)
The distance between two points, 'A', and 'B', on the coordinate plane given their coordinates, (x₁, y₁), and (x₂, y₂) can be found using following formula;
Substituting the known 'x', and 'y', values for the coordinates of the points, we have;
Therefore, the distance between the points, (2, -2), and (6, 3) = √(41) ≈ 6.4.
You will save $9
Convert 20% to decimal
Each % is 0.01 so 20×0.01=0.2
Then multiply
45×0.2=9
Step-by-step explanation:
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<em><u>concept :</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2>
<em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2>
<em><u>5y + 4x = 35</u></em></h2><h2 /><h2>
<em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2>
<em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2>
<em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2>
<em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2>
<em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2>
<em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>
