Answer:
solution: KL-2ML. LKIN=63x2=126*
Step-by-step explanation:
Mark me brainliest!!
Answer:
I can answer the question, but represent in what? Do u got a picture or example or sum?
Answer:
log 5
Step-by-step explanation:
We can rewrite this by using a property of logs
log a - log b = log (a/b)
log(10) - log(2)
log(10/2)
log 5
Problem 1, part (a)
<h3>Answer: False</h3>
For instance, 200 feet in real life can be reduced to scale down to say 2 inches on paper. So we have a reduction going on, and not an enlargement.
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Problem 1, part (b)
<h3>Answer: true</h3>
This is because a scale drawing involves similar polygons. This is true whenever any dilation is applied.
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Problem 2
I'm not sure how your teacher wanted you to answer this question. S/he didn't give you any numbers for the side lengths of the polygon. The angle measures are missing as well.
The length and width that will maximize the area are 175 ft and 87.5 ft respectively
The largest area that can be enclosed is 15312.5 ft²
<h3>Area of a rectangle</h3>
where
l = length
w = width
The fencing is 350 ft it is use to enclose a rectangular plot with a river occupying one part.
Therefore,
perimeter = l + 2w
350 = l + 2w
l = 350 - 2w
area = (350 - 2w)w
(350 - 2w)w = 0
where
w = 0 or 175
average = 175/2 = 87.5
Hence, the max area is at w = 87.5 ft
Therefore,
l = 350 - 2(87.5) = 175 ft
length = 175 ft
width = 87.5 ft
Therefore,
area = 175 × 87.5 = 15312.5 ft²
Therefore, the largest area that can be enclosed is 15312.5 ft²
learn more on rectangle here: brainly.com/question/11630499
