A chemist is using 362 milliliters of a solution of acid and water. If 12.1% of the solution is acid, how many milliliters of ac
2 answers:
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Now, bearing in mind that, for a graph, to be "differentiable", means a "smooth transition", not an abrupt edge or a "cusp", the first subfunction of the piece-wise is just a quadratic, the second subfunction is a linear, well, so the linear needs to hit the quadratic "smoothly", not make an abrupt edge, for that, the linear must continue the quadratic from 2 onwards, since 2 is the range edge.
now, what is the piece-wise when x = 2? well, f(2) = x² or (2)² or 4
that means, the linear needs to hit it at that point, to make it so, let's make it the slope of the linear, the same as the quadratic's.
what's the quadratic's slope? well, simple enough
so we have y = 4x + b
now, recall, for the quadratic, when x =2, y = 4, so for the linear, when x = 2, y = 4 as well, this means
check the picture below.
now, what is the piece-wise when x = 2? well, f(2) = x² or (2)² or 4
that means, the linear needs to hit it at that point, to make it so, let's make it the slope of the linear, the same as the quadratic's.
what's the quadratic's slope? well, simple enough
so we have y = 4x + b
now, recall, for the quadratic, when x =2, y = 4, so for the linear, when x = 2, y = 4 as well, this means
check the picture below.
Answer:
Segment EF: y = -x + 8
Segment BC: y = -x + 2
Step-by-step explanation:
Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.
Upon observing the given diagram, we can infer the following corresponding sides:
We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.
<h2>Slope of Segment BC:</h2>In order to solve for the slope of segment BC, we can use the following slope formula:
Use the following coordinates from the given diagram:
Point B: (x₁, y₁) = (-2, 4)
Point C: (x₂, y₂) = ( 1, 1 )
Substitute these values into the slope formula:
Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:
Point E: (x₁, y₁) = (4, 4)
Point F: (x₂, y₂) = (6, 2)
Substitute these values into the slope formula:
Our calculations show that segment BC and EF have the same slope of -1. In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.
Since segments BC and EF have the same slope, then it means that .
The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.
Using the slope of segment BC, m = -1, and the coordinates of point C, (1, 1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>
y = mx + b
1 = -1( 1 ) + b
1 = -1 + b
Add 1 to both sides to isolate b:
1 + 1 = -1 + 1 + b
2 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.
Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:
⇒ y = -x + 2.
<h3><u /></h3><h3><u>Segment EF:</u></h3>Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>
y = mx + b
4 = -1( 4 ) + b
4 = -4 + b
Add 4 to both sides to isolate b:
4 + 4 = -4 + 4 + b
8 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.
Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:
⇒ y = -x + 8.