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Law Incorporation [45]
3 years ago
13

a plant nursery is growing a tree that is 3 ft tall and grows at an average rate of 1 ft per year. Another tree at the nursery i

s 4 ft tall and grows at an average rate of 0.5 ft per year. After how many years will the trees be the same height?
Mathematics
1 answer:
Sav [38]3 years ago
8 0
After 2 years. because the first tree will grow 2ft and the second will grow 1ft, which will make them both 5 feet tall
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Line segment YZ was used to translate ABCDE. YZ is 6.2 inches long. What is the length of AA'+BB'+CC'+DD'+EE'?
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The length is 31 inches
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3 years ago
Find the surface area of the equilateral triangular prism below.
velikii [3]

Answer:

Area of a triangle = bh/2

Area of a rectangle = bh

The area of the triangles are the same, the area of the rectangles are the same.

triangle: 9x8/2

72/2 = 36

Rectangle: 8x14=112

There are 2 triangles and 3 rectangles so,

2(36)+3(112)

72+336= 408 in²

3 0
3 years ago
In a fruit cocktail, for every 15 ml of orange juice you need 25 ml of apple juice and 10 ml of coconut milk. What proportion of
frosja888 [35]

Answer:

3/10 is orange juice.

Step-by-step explanation:

Step 1: Find the total mL in the concoction

15 + 25 + 10 = 50 mL in a fruit cocktail

Step 2: Find the proportion of orange juice in 1 serving (50 mL) of cocktail

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8 0
4 years ago
HELP ON GEOMETRY!!! PHOTO ATTACHED
Dvinal [7]

Answer:

q = 3 units

Step-by-step explanation:

In\: \triangle GFH \: and\; \triangle GJI\\\angle GFH \cong \angle GJI..(each\: 90\degree)\\\angle FGH \cong\angle JGH..(common\: angles)\\\therefore \triangle GFH \sim \triangle GJI..(by \: AA\: postulate)\\\\\therefore \frac{GF}{GJ}=\frac{HF}{IJ}..(By\: c.s.s.t.)\\\\\therefore \frac{4}{8}= \frac{q}{6}\\\\\therefore \frac{1}{2}= \frac{q}{6}\\\\q =\frac{6}{2}\\q= 3 \: units

7 0
3 years ago
Premises: All good students are good readers. Some math students are good students. Conclusion: Some math students are good read
just olya [345]

Answer:

Therefore, the conclusion is valid.

The required diagram is shown below:

Step-by-step explanation:

Consider the provided statement.

Premises: All good students are good readers. Some math students are good students.

Conclusion: Some math students are good readers.

It is given that All good students are good readers, that means all good students are the subset of good readers.

Now, it is given that some math students are good students, that means there exist some math student who are good students as well as good reader.

Therefore, the conclusion is valid.

The required diagram is shown below:

3 0
3 years ago
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