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vekshin1
3 years ago
6

If a 24 foot tree casts a 7 foot shadow then what is the length of the shadow that 4 foot tall girl casts? Choose best possible

answer.
1. 28 feet
2. 5.8 feet
3. 1.2 feet
4. None
Mathematics
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

hmmm

Step-by-step explanation:

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Solve for x.<br> 21x + 11 - 5 = 5
Reil [10]

Answer:

x= -1/21

Step-by-step explanation:

21x + 11 - 5 = 5

21x + 6 = 5

(-6 each side)

21x = -1

(divide 21 each side)

x= -1/21

3 0
3 years ago
Read 2 more answers
I’m confused on this one
kvv77 [185]
The answer is 42.

The question is asking for angle N. We know from the question that angle N is equal to angle E (Don't think about the scale factor because it only applies to the sides. They're just trying to trick you).

To find angle E:
79 + 6x + 11  + 7x - 14 = 180
Move numbers to the right side:
6x + 7x = 180 - 79 - 11 + 14
Combine like terms:
13x = 104
Divide both sides by 13:
x = 8
The formula for angle E:
7x - 14
Plug in the 8:
7 \times 8 - 14 = 56 - 14 = 42
The angle measure of E is 42. So angle N is automatically 42 too.

To solidify, we can try to use the formula which they give us for angle N:
3x + 18 = 24 + 18 = 42
8 0
2 years ago
Find BC<br> (triangle)<br> sides = 20mi, 22mi<br> inside = 95 degrees
Hunter-Best [27]
To solve this problem you need to know the law of cosines.

7 0
3 years ago
Write 87/1000 as a decimal.
Akimi4 [234]
87/1000 into a decimal is 0.087,
You can divide with your calculator, or you can use place value.
0.8 = 8/10
0.87 = 87/100
0.087 = 87/1000
And so on and so forth...
5 0
3 years ago
Read 2 more answers
The data represents the body mass index? (bmi) values for 20 females. construct a frequency distribution beginning with a lower
Reil [10]
Given the data below <span>representing the body mass index (bmi) values for 20 females.
17.7     33.5     26.9     22.7     22.2     29.9     23.6     18.3     27.7     23.4     19.2     25.9     22.9     37.7     31.6     28.1     44.9     31.6     25.2     23.9

From the data above, we construct a frequency distribution beginning with a lower class limit of 15.0 and use a class width of 6.0 as follows:

Class iinterval: 15.0 - 21.0   21.0 - 27.0   27.0 - 33.0   33.0 - 39.0   39.0 - 45.0
Frequency:              3                  9                  5                   2                  1

From the frequency table, it can be seen that the frequencies started low, increased to a point and then decrease. It can also be seen that the highest frequency of the data is not at the center of the distribution, so the distribution is not symetric.

Therefore, the frequency distribution does not appear to be roughly a normal distribution, because, "although the frequency start low, increase to some maximum, then decrease, the distribution is not symmetric."</span>
5 0
3 years ago
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