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lorasvet [3.4K]

3 years ago

14

Please help me find the solution

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