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Papessa [141]
3 years ago
11

Will mark BRAINLIEST!!!!

Mathematics
1 answer:
Virty [35]3 years ago
6 0

Answer:

The first one is your answer.

Step-by-step explanation:

What you do is you have to look at the 2/18 in the square root. That can come out to be 1/3. So you know that a 3 is going to be on the bottom.

Now you look at the x^5. What you do is you make it until you have 5 x's like this x, x, x, x, x. You then put them in pairs of two.

x,x   x,x   x.  There are two pairs of two so that comes out.

So you have 1x^2sqrtx/3. which is the first one.

#1 is your answer.

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Find the fifth roots of 243(cos 240° + i sin 240°). (5 points)
MakcuM [25]

Answer:3

Step-by-step explanation:r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240), then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).

3 0
3 years ago
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5 0
3 years ago
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Find two consecutive integers whose product is 50
GarryVolchara [31]

n, n+1 - two consecutive integers

n(n + 1) = 50     <em>use distributive property</em>

n² + n = 50     <em>subtract 50 from both sides</em>

n² + n - 50 = 0

-----------------------------------------------------

ax² + bx + c =0

if b² - 4ac > 0 then we have two solutions:

[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a

if b² - 4ac = 0 then we have one solution -b/2a

if  b² - 4ac < 0 then no real solution

----------------------------------------------------------

n² + n - 50 = 0

a = 1, b = 1, c = -50

b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions

√(b² - 4ac) = √(201) - it's the irrational number

Answer: There are no two consecutive integers whose product is 50.

3 0
4 years ago
Pls help in this<br> 7.6x5.2
Nimfa-mama [501]

Answer:

39.52

Step-by-step explanation:

3 0
3 years ago
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Answer:

A.

Step-by-step explanation:

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