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yanalaym [24]
3 years ago
8

Which of the following is the solution set of x2 + 8x + 12 = 0? (-2.-6 2.6

Mathematics
1 answer:
algol [13]3 years ago
6 0

To solve this polynomial equation, we first factor the left side.

To factor x² + 8x + 12 = 0, look for factors of 12 that add to 8.

These factors are 6 and 2 so we have (x + 6)(x + 2) = 0.

So either x + 6 = 0 or x + 2 = 0 and solving for

x in each equation, our answer is {-6, -2}.

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0.4= Blankx10 please answer quick
Marta_Voda [28]

Answer:

1, 0.4

2, 0.004

3, 0.04

Step-by-step explanation:

hope this helps.      :)

6 0
2 years ago
I need help on questions two lol pls help me
aksik [14]
1. (8/9)/4

8/9 * 1/4 = 8/36= 2/9 paint for each birdhouse

2. ron 1 mile every 20 minutes so 3 mph

1 mile every 24 minutes so ron is the faster walker
5 0
2 years ago
Hey guys please help me out. Math isn't exactly my strong subject and I'd really appreciate this answer with steps. Thank you in
mash [69]
B1 = 2
b2 = (b1)^2 + 1 = 2^2 + 1 = 5
b3 = (b2)^2 + 1 = 5^2 + 1 = 26

b4 = (b3)^2 + 1 = 26^2 + 1 = 676+1=<span>677</span>
8 0
2 years ago
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A tetrahedron has 4 sides. If the numbers 1, 2, 3, and 4 are on the faces of the tetrahedron, what
Savatey [412]

Probability of randomly rolling a 3 is 1/4

Step-by-step explanation:

Step 1:

It is given that a tetrahedron has 4 sides with 1,2,3 and 4 and is randomly rolled. We need to compute the probability of randomly rolling a 3.

Step 2:

Probability of an event = (Favorable outcomes) / (Total outcomes).

There are 4 total outcomes when the tetrahedron is rolled, these are the values 1 to 4.

The only favorable outcome is getting a 3

Probability of getting a 3 when the tetrahedron is randomly rolled = 1/4

Step 3:

Answer:

Probability of randomly rolling a 3 is 1/4

7 0
3 years ago
100 points! simplify write as a product compute
Rom4ik [11]

Answer:

a) \sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

b)( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) \frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =   - 1

Step-by-step explanation:

We want to simplify

\sqrt{61 - 24 \sqrt{5} }

Let :

\sqrt{61 - 24 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation:

(\sqrt{61 - 24 \sqrt{5} } )^{2}  =  ({a - b \sqrt{5} })^{2}

Expand the RHS;

61 - 24 \sqrt{5} =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients on both sides:

{a}^{2}  + 5 {b}^{2}  = 61 -  -  - (1)

- 24 =  - 2ab \\ ab = 12 \\ b =  \frac{12}{b}  -  -  -( 2)

Solve the equations simultaneously,

\frac{144}{ {b}^{2} }  + 5 {b}^{2}  = 61

5 {b}^{4}  - 61 {b}^{2}  + 144 = 0

Solve the quadratic equation in b²

{b}^{2}  = 9 \: or \:  {b}^{2}  =  \frac{16}{5}

This implies that:

b =  \pm3 \: or \: b =  \pm  \frac{4 \sqrt{5} }{5}

When b=-3,

a =  - 4

Therefore

\sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

We want to rewrite as a product:

{b}^{2}  {c}^{2}  - 4bc -  {b}^{2}  -  {c}^{2}  + 1

as a product:

We rearrange to get:

{b}^{2}  {c}^{2}   -  {b}^{2}  -  {c}^{2}  + 1- 4bc

We factor to get:

{b}^{2} ( {c}^{2}   -  1)  -  ({c}^{2}   -  1)- 4bc

Factor again to get;

( {c}^{2}   -  1) ({b}^{2}   -  1)- 4bc

We rewrite as difference of two squares:

(\sqrt{( {c}^{2}   -  1) ({b}^{2}   -  1) })^{2} - ( {2 \sqrt{bc} })^{2}

We factor the difference of square further to get;

( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) We want to compute:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }

Let the numerator,

\sqrt{9 - 4 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation;

9 - 4 \sqrt{5}  =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients in both equations;

{a}^{2}  + 5 {b}^{2}  = 9 -  -  - (1)

and

- 2ab =  - 4 \\ ab = 2 \\ a =  \frac{2}{b}  -  -  -  - (2)

Put equation (2) in (1) and solve;

\frac{4}{ {b}^{2} }  + 5 {b}^{2}  = 9

5 {b}^{4}   - 9 {b}^{2}  + 4 = 0

b =  \pm1

When b=-1, a=-2

This means that:

\sqrt{9 - 4 \sqrt{5} }  =  - 2 +  \sqrt{5}

This implies that:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =  \frac{ - 2 +  \sqrt{5} }{2 -  \sqrt{5} }  =  \frac{ - (2 -  \sqrt{5)} }{2 -  \sqrt{5} }  =  - 1

3 0
3 years ago
Read 2 more answers
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