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mihalych1998 [28]

2 years ago

7

Algebra help - please be so kind - god bless. 2&3 please

1 answer:

azamat2 years ago

4 0

2. -11 sq rt of 112
Sq rt of 16 times sq rt of 7
Becomes 4 sq rt of 7
Multiply -11 times 4 sq rt of 7
-44 sq rt of 7
Answer A

3. Subtract 17-9 since they are hitch the same sq rt

Answer 8 sq rt of 17

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Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\$

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$

<u>Separate the solutions</u>

$k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2}$

The equation has complex roots

<u>Separate the real and imaginary parts</u>

We have the final answer as

$k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i$

Hope this helps you

8 0

2 years ago

I need help, an answer or how to- will be fine, thank you. (i’ll mark brainliest if i can)

siniylev [52]

by solving the first equation you'll get x= -24/13

if you solve the second option (B) then you'll the same result i.e x=-24/13

so option C is correct

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2 years ago

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Elena L [17]

Answer:

prime

Step-by-step explanation:

3x^2 + 5x + 1

it can't be factorizes with rational numbers

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