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Greeley [361]
3 years ago
8

Don't know what to do! Helllllp plz! Probability!!!!!

Mathematics
2 answers:
In-s [12.5K]3 years ago
7 0

Answer:

6 passwords

Step-by-step explanation:

1. 718

2. 781

3. 817

4. 871

5. 178

6. 187

Westkost [7]3 years ago
4 0
6 ways

3! = 6

sorry if this is wrong haven’t done this in a minute
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mel-nik [20]

Answer:

y=2/3x-8

Step-by-step explanation:

2x-3y=24

3y=2x-24

y=2/3x-24/3

y=2/3x-8

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3 0
2 years ago
What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*
pickupchik [31]
<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation x^4-41x^2=-400.

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

x^4-41x^2+400=-400+400.

x^4-41x^2+400=0.

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

x^4-25x^2-16x^2+400=0

Making it into two groups, we get

(x^4-25x^2)+(-16x^2+400)=0

Factoring out GCF of each group, we get

x^2(x^2-25)-16(x^2-25)=0

(x^2-25)(x^2-16) =0

Factoring out  (x^2-25) and (x^2-16) separately by difference of the squares identity a^2-b^2=(a-b)(a+b), we get

(x^2-25) = x^2-5^2= (x-5)(x+5) and

x^2-16 = x^2-4^2 =(x-4)(x+4).

Therefore,

(x-5)(x+5)(x-4)(x+4) =0

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>
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Answer:

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Step-by-step explanation:

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