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belka [17]
4 years ago
15

I need help on my ixl

Mathematics
1 answer:
mojhsa [17]4 years ago
7 0
Area = W * L55 = 5 * L55/5 = L11 = L
Note "L" is the length.
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ABC
aev [14]

Answer:

A

Step-by-step explanation:

If 50 is the diameter that thr radius is 25 and the equation to find the area of a circle is pi(r^2). So you would do pi(25^2) and that is 1963

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What is y=7/4x+6 if the input was 53
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7/4(53) + 6
92.75 + 6

98.75
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The graph of the function y=f(x)
jarptica [38.1K]
The range is the high and low points. 

So (5, 2] 
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The system of equations Y =-3x+2 and Y = 3x-6 is shown on the graph below.
Ira Lisetskai [31]

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8 0
3 years ago
What is the tan invers of 3i/-1-i​
postnew [5]

<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> )

<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> ) × (-1 + <em>i</em> ) / (-1 + <em>i</em> )

<em>z</em> = (3<em>i</em> × (-1 + <em>i</em> )) / ((-1)² - <em>i</em> ²)

<em>z</em> = (-3<em>i</em> + 3<em>i</em> ²) / ((-1)² - <em>i</em> ²)

<em>z</em> = (-3 - 3<em>i </em>) / (1 - (-1))

<em>z</em> = (-3 - 3<em>i </em>) / 2

Note that this number lies in the third quadrant of the complex plane, where both Re(<em>z</em>) and Im(<em>z</em>) are negative. But arctan only returns angles between -<em>π</em>/2 and <em>π</em>/2. So we have

arg(<em>z</em>) = arctan((-3/2)/(-3/2)) - <em>π</em>

arg(<em>z</em>) = arctan(1) - <em>π</em>

arg(<em>z</em>) = <em>π</em>/4 - <em>π</em>

arg(<em>z</em>) = -3<em>π</em>/4

where I'm taking arg(<em>z</em>) to have a range of -<em>π</em> < arg(<em>z</em>) ≤ <em>π</em>.

6 0
3 years ago
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