All categories

3 years ago

K^2+5k-6=0 quadratic formula

1 answer:

s2008m [1.1K]3 years ago

7 0

I hope this helps you

1.k^2+5k-6=0

a=1

b=5

c= -6

disctirminant =b^2-4ac

disctirminant =5^2-4.1. (-6)

disctirminant =49

x1= -(5)+ square root of 49/2.1

x1=-5+7/2

x1=1

x2= -5- square root of 49/2.1

x2= -5-7/2

x2= -6

You might be interested in

X + 3/ 5 = 2 -5 -3 7 13

ahrayia [7]

X 3/5 =2x+15=2x=15-2x=13

answer 13

4 0

2 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS

ladessa [460]

Answer:

-8 vertical in negative one horizontal

Step-by-step explanation:

Mark me brainliest!!

6 0

2 years ago

Read 2 more answers

I need help please and thank you

sergeinik [125]

Try googling it , sorry

8 0

2 years ago

Help me please!!!!!!!!!!!

frozen [14]

Answer:

Step-by-step explanation:

Socratic had the answer on there I think

7 0

3 years ago

What is the cube root of 27a^l2? -3a4 -3a 3a 3a*

GuDViN [60]

Answer:

$3a^4$

Step-by-step explanation:

What is the cube root of $27a^{12}$ ? This is the question.

We can write:

$\sqrt[3]{27a^{12}}$

We will use the below property to simplify:

$\sqrt[n]{a*b}=\sqrt[n]{a} \sqrt[n]{b}$

So, we have:

$\sqrt[3]{27a^{12}} =\sqrt[3]{27} \sqrt[3]{a^{12}}$

We will now use below property to further simplify:

$\sqrt[n]{x} =x^{\frac{1}{n}}$

Thus, we have:

$\sqrt[3]{27} \sqrt[3]{a^{12}} =3*(a^{12})^{\frac{1}{3}}$

We know power to the power rule: $(a^z)^b=a^{zb}$

Now, we have:

$3*(a^{12})^{\frac{1}{3}}\\=3*a^{\frac{12}{3}}\\=3a^4$

This is the correct answer: $3a^4$

4 0

3 years ago