If Dorie devotes the same number of hours to studying for these three classes each week, what is the total number of hours Kathl
een will study over a three week period?
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The original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m
<h3>How does dilation affect length, area, and volume of an object?</h3>Suppose a figure (pre image) is dilated (dilated image) by scale factor of k.
So, if a side of the figure is of length L units, and that of its similar figure is of M units, then:
<u><em>L = k × M</em></u>
where 'k' will be called as scale factor.
The linear things grow linearly like length, height etc.
The quantities which are squares or multiple of linear things twice grow by square of scale factor. Thus, we need to multiply or divide by <u><em>k²</em></u><em> </em>to get each other corresponding quantity from their similar figures' quantities.
<u><em>So </em></u><u><em>area </em></u><u><em>of </em></u><u><em>first figure </em></u><u><em>= k² × area of second figure</em></u>
Similarly,
<u><em>Volume </em></u><u><em>of first figure = k³ × volume of second figure.</em></u>
It is because we will need to multiply 3 linear quantities to get volume, which results in k getting multiplied 3 times, thus, cubed.
For this case, we're given that;
- Surface area of pre-image = 35 m²
- Surface area of dilated image = 315 m²
- Height of pre-image = 5.8 m
- Height of dilated image.
Let the scale fator of dilation be k.
Then:
<u><em>So area of pre-image = k² × area of dilated image</em></u>
Thus, we get:
(took positive root as scaling is done by non-negative factors as only magnitude matters mostly)
Thus, we get:
Height of pre-image = (1/3) × Height of dilated image (say h)
Thus, the original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m
Learn more about dilation here:
Answer:
x - 8y - z = 1
Step-by-step explanation:
Data provided according to the question is as follows
f(x,y) = z = ln(x - 8y)
Now the equation for the tangent plane to the surface
For z = f (x,y) at the point P is
Now the partial derivatives of f are
= 1
Now
= -8
So, the tangent equation is
Now after solving this, the following equation arise
z = x - 9 - 8y + 8
z = x - 8y - 1
Therefore
x - 8y - z = 1