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lorasvet [3.4K]
3 years ago
14

Please help me find the solution ​

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
C.) (-4,3)
Use the process of elimination and plug in the answer of x or y to one of the problems

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Write the complex number in the form a + bi.
bonufazy [111]

Answer:

\frac{3}{2} + \frac{3\sqrt{3} }{2} i

Step-by-step explanation:

Using the exact value

cos60° = \frac{1}{2} and sin60° = \frac{\sqrt{3} }{2}, then

3(cos60° + isin60°)

= 3( \frac{1}{2} + \frac{\sqrt{3} }{2} i) ← distribute by 3

= \frac{3}{2} + \frac{3\sqrt{3} }{2} i ← in the form a + bi

7 0
3 years ago
What is three and seven hundredths standard form
Valentin [98]

Answer: I believe the answer is 3.07 but I may be wrong so if you want you can wait for another answer :D

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Help me guys <br> thx so much
navik [9.2K]

Answer:

Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Step-by-step explanation:

<u>Step 1:  Multiply</u>

<u />(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\

(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}

(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x}  )

2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}

10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Answer:  Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

6 0
3 years ago
Read 2 more answers
A bourse travels 270 miles in 30 hours what is it’s speed
Likurg_2 [28]

Answer:

9 miles per hour

Step-by-step explanation:

6 0
3 years ago
Hello please help with my question, I will give brainlyest!
lilavasa [31]

No, John is incorrect.

<h3>Correct work shown:</h3>

\sf x^2 - 6x - 7 = 0

\sf x^2 - 6x = 7

\sf x^2 - 6x  + 9= 7 + 9

\sf (x-3)^2= 16

\sf x-3=\pm \sqrt{16}

\sf x-3= \pm 4

\sf x= + 4 + 3 \quad or \quad x = -4 + 3

\sf x= 7 \quad or \quad x = -1

The correct answer should be x = 7 or x = -1

3 0
1 year ago
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