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romanna [79]
3 years ago
6

It’s not a or c help

Mathematics
2 answers:
Luden [163]3 years ago
6 0

Answer:

D if i'm wrong i'm so sorry

Step-by-step explanation:

svp [43]3 years ago
3 0
It is b i worked it out and got b
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What percentage of 40 is 12?
dimulka [17.4K]

Answer:

12 is 3 percent of 40.

Step-by-step explanation:

If you divide 12 by 40, you get 0.3. 0.3 as a fraction is 3/100. So, it would be 3% of 100.

6 0
3 years ago
Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.
Gnesinka [82]

Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, \overline{CP} ⊥ \overline{AM}, AP = 9, and PM = 16

\overline{AC}² + \overline{CM}² = \overline{AM}²

\overline{AM} = \overline{AP} + PM = 9 + 16 = 25

\overline{AM} = 25

\overline{AC}² = \overline{AP}² + \overline{CP}² = 9² +  \overline{CP}²

∴ \overline{AC}² = 9² +  \overline{CP}²

Similarly we get;

\overline{CM}² = 16² + \overline{CP}²

Therefore, we get;

\overline{AC}² + \overline{CM}² = 9² +  \overline{CP}² + 16² + \overline{CP}² = \overline{AM}² = 25²

2·\overline{CP}² = 25² - (9² + 16²) = 288

\overline{CP}² = 288/2 = 144

\overline{CP} = √144 = 12

From \overline{AC}² = 9² +  \overline{CP}², we get

\overline{AC} = √(9² +  12²) = 15

\overline{AC} = 15

From, \overline{CM}² = 16² + \overline{CP}², we get;

\overline{CM} = √(16² + 12²) = 20

\overline{CM} = 20.

3 0
2 years ago
Surface Area of Cylinders
lidiya [134]

Answer: 1639.08

Step-by-step explanation: To find the surface area of a cylinder, multiply 2(3.14)times the radius squared. Your answer is 508.68. Then multiply 2(3.14)times the radius and the height. You'll get 1130.08. add the answers you got from both equations together. That's the surface area of your cylinder. If you have any more questions, please let me know.

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3 years ago
What is 1/6 + 1/4 + 1/3
IgorC [24]
The answer is 9/12

Brainliest pls
6 0
3 years ago
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What is the area of the trapezoid shown below?​
anzhelika [568]

Answer:

78 units squared

Step-by-step explanation:

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2 years ago
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