Measurement of one angle and length of one side I think
The length and width of the garden will be obtain as follows:
let the length be x ft, the width will be (x+3.5)ft
area of the bed will be:
A=length*width
11.76=x(x+3.5)
11.76=x^2+3.5x
thus
x^2+3.5x-11.76=0
solving this by quadratic formula we get:
x=[-b+/-sqrt(b^2-4ac)]/2a
thus plugging in the values we obtain
x=[-3.5+/-sqrt(3.5^2-4(-11.76*1))]/2
x=[-3.5+/-sqrt59.29]/2
x=[-3.5+/-7.7]/2
x=4.2 or -11.2
thus the length will be x=4.2 ft and the width will be x-3.5=0.7 ft
Answer: B) Dilate by scale factor of 2
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Explanation:
Your teacher isn't saying this directly, but I'm assuming s/he wants you to find a similar figure that isn't congruent to the original. Informally, your teacher seems to want you to find a figure that is the same shape but not the same size as the original.
If so, then any dilation will shrink or enlarge the image depending on the scale factor. So the new image will not be the same as the old one. In this case, a dilation with scale factor 2 means the new figure is twice as large (each side is twice as long). But the old image is similar to the new image. The angles keep their values and therefore we get the same shape. This is why choice B is the answer. Again this is assuming what I mentioned in the first paragraph.
Choices A, C, and D are all known as rigid transformations and they preserve the same size of the figure. Applying any of those operations will lead to the same figure (just rotated, reflected or shifted somehow). In other words, applying operations A,C, or D will have us get two congruent triangles. If two triangles are congruent, then they are automatically similar, but not vice versa. This is why we can rule out A,C, and D.
As is the case for any polynomial, the domain of this one is (-infinity, +infinity).
To find the range, we need to determine the minimum value that f(x) can have. The coefficients here are a=2, b=6 and c = 2,
The x-coordinate of the vertex is x = -b/(2a), which here is x = -6/4 = -3/2.
Evaluate the function at x = 3/2 to find the y-coordinate of the vertex, which is also the smallest value the function can take on. That happens to be y = -5/2, so the range is [-5/2, infinity).