Answer:m=2k/v^2
Step-by-step explanation:
The first thing you should do is divide 1/2 on both sides. Now you have 2k=mv^2. Then divide v^2 from each side and you get a final answer of (2k/v^2)=m
Point (2, 4) was reflected over the x axis to give (2, -4). It was then dilated by a scale factor of 2 to get (4, -8). Hence (2, 4) ⇒ (4, -8)
<h3>What is a
transformation?</h3>
Transformation is the movement of a point from its initial location to a new location. Types of transformation are <em>reflection, translation, rotation and dilation.</em>
Translation is the movement of a point either<em> up, down, left or right</em> on the coordinate plane.
Point (2, 4) was reflected over the x axis to give (2, -4). It was then dilated by a scale factor of 2 to get (4, -8). Hence (2, 4) ⇒ (4, -8)
Find out more on transformation at: brainly.com/question/4289712
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1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
-2/3x+6
Step-by-step explanation:
Answer:
<em>y = (-4/3)*x + 7</em>
<em />
Step-by-step explanation:
The point-slope form of the equation of a line is: <em>y = a*x + b </em>
In the above equation, <em>a </em>is the slope of the line representing the equation in the graph and y is the function of x [y = y(x)]
The given line has a slope of -4/3, so that <em>a = -4/3 </em>
=> The equation of this line has a form as following: <em>y = (-4/3)*x + b (1)</em>
<em />
As the line passes through the point (9; -5) (in which: x = 9; y = -5). Replace x =9 and y = -5 into the equation (1), we have:
<em>y = (-4/3)*x + b</em>
<em>=> -5 = (-4/3)*9 + b </em>
<em>=> -5 = -12 + b </em>
<em>=> b = 7</em>
<em />
So that the equation in point-slope form of the given line is:
<em>y = (-4/3)*x + 7</em>
<em />
