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hram777 [196]
3 years ago
8

LOTS OF POINTS please help with three questions

Mathematics
2 answers:
Law Incorporation [45]3 years ago
8 0

Answer:

above is correct

Step-by-step explanation:

ser-zykov [4K]3 years ago
6 0

Answer:

1. 40

2. 91

idk 3... Sorry..

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What plus what equals 98
otez555 [7]
49+49=98

do you need more answers than that?? cause there are almost an infinite range of what can equal 98 when added together..
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3 years ago
Can someone please help me answer theses questions <br><br> Thank u so much
aleksandr82 [10.1K]
A polynomial with 3 terms:
x^2 +3xyz -4

It meets the definition of a polynomial because it is a sum of products or constants, and all variables have non-negative integer powers.
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(sqrt3-sqrt3i)^4
Ludmilka [50]

The increasing order of the complex numbers is (√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

<h3>Absolute values of the complex numbers</h3>

The absolute values of the complex numbers are determined as follows;

(sqrt3-sqrt3i)^4 = (√3 - √3i)⁴

|z| = \sqrt{(\sqrt{3} )^2 + (\sqrt{3 }\times1 )^2} } \\\\|z| = \sqrt{6}

(-1+sqrt3i)^12 = (-1 + √3i)¹²

|z| = \sqrt{(-1)^2 + (\sqrt{3)^2} } \\\\|z| = \sqrt{4} \\\\|z| = 2

(sqrt 3-i)^6 = (√3 - i)⁶

|z| = \sqrt{(\sqrt{3})^2 + (-1)^2 } \\\\|z| = \sqrt{4} \\\\|z| = 2

(sqrt2-sqrt2i)^8 = (√2 - √2i)⁸

|z| = \sqrt{(\sqrt{2} )^2 + (\sqrt{2})^2 } \\\\|z| = 2

(sqrt2-i)^6 = (√2 - i)⁶

|z| = \sqrt{(\sqrt{2})^2 + (-1)^2} } \\\\|z| = \sqrt{3}

Increasing order of the complex numbers;

(√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

Learn more about complex numbers here: brainly.com/question/10662770

#SPJ1

3 0
3 years ago
What is the definition of indemnity?
Gemiola [76]
<span>security or protection against a loss or other financial burden.
</span>
4 0
3 years ago
Read 2 more answers
Help i cant figure this out:(
stealth61 [152]

Answer:

6-2=4

31/5 x 8 = 248/40

15/8 x 5 = 75/40

248-75=173

4 13/40

answers in order

6-2=4

6 1/5 - 1 7/8 = 4 13/40

Hope this helps

Step-by-step explanation:

8 0
3 years ago
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