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Sever21 [200]
3 years ago
11

How do you write 57 and 62/100 in expanded form

Mathematics
2 answers:
aksik [14]3 years ago
7 0
57 divided by 100 and 62 divided by 100
AlekseyPX3 years ago
7 0
57.62

50+7+6/10+2/100=57.62
-------------------------------------
57

50+7=57
-------------------------------------
62/100

6/10+2/100=62/1000


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Answer:

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The base of a solid in the first quadrant of the xy plane is a right triangle bounded by the coordinate axes and the line x + y
Ivahew [28]

The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is 

(x - x^2)^2 * sqrt(3)/4. 

Integrating from x = 0 to x = 1, we have 

[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4 

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Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...

8 0
3 years ago
Write an equation of the line that passes through the points.<br><br> (0,1),(−2,−5)
Arada [10]

Answer:

y=4/-2

Step-by-step explanation:

I think that is the answer

6 0
2 years ago
Read 2 more answers
What is the number of the variable X?​
Olenka [21]

Answer:

\huge\boxed{x = -18}

Step-by-step explanation:

\frac{x-3}{3} = \frac{2x+1}{5} \\\\Cross \ Multiplying\\\\3(2x+1) = 5(x-3) \\\\Resolving \ Parenthesis\\\\6x+3 = 5x-15\\\\Combining \ like \ terms\\\\6x-5x = -15-3\\\\\bold{x = -18}

Hope this helped!

<h2>~AnonymousHelper1807</h2>
7 0
2 years ago
What is the probability that the number of “HEADs” on these four coins is equal to 3? (4 points)
slavikrds [6]

Answer:

a. \frac{1}{4}  

Step-by-step explanation:  

We are asked to find the probability of getting 3 heads on 4 flips.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

Since we know that flipping a fair coin has 2 equally likely possible outcomes, so flipping four coins will have 2*2*2*2=16 possible outcomes.

Sample space of possible outcomes.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH,THTT, TTHH, TTHT, TTTH, TTTT.

We can see that there are 4 favorable outcomes of getting heads.

\text{Probability of getting 3 heads}=\frac{4}{16}

\text{Probability of getting 3 heads}=\frac{1}{4}

Therefore, the probability of getting 3 heads on 4 coins will be \frac{1}{4} and option a is the correct choice.

5 0
3 years ago
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