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2 years ago

3x2 - 5 = -8x – 10 HELP PLEASE

Mathematics

1 answer:

Travka [436]2 years ago

5 0

3

2
−
5
=
−
8

−
1
0
3

2
−
5
−
(
−
8

−
1
0
)
=
0

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REPOST*

xeze [42]

-6-3-5

Hence inequality will be

$\\ \sf\longmapsto -6\leqslant 5$

#2.

But we need inequality

$\\ \sf\longmapsto 3\leqslant 5$

The inequality given by

$\\ \sf\longmapsto -6\leqslant 3\leqslant 5$

7 0

2 years ago

Read 2 more answers

10 = 3x – 8 what is this

Margaret [11]

Answer:

Answer is below

Step-by-step explanation:

10=3x-8

always remember to flip the equation

3x-8=10

Add 8 to both sides

3x-8(+8)=10(+8)

3x=18

Divide both sides by 3

3x/3=18/3

x=6

8 0

3 years ago

Read 2 more answers

What’s the scientific notation for 3,207,000,000

Pani-rosa [81]

Answer:

3.207 x $10^{9}$

Step-by-step explanation:

8 0

3 years ago

A number cube has the numbers 1, 2, 5, 5, 6, and 6 on its faces. Match the probability with the outcome. 1. P(5 ∪ 6) 2. P(2 ∪ 5

Tems11 [23]

Answer:

1. P(5 U 6) =2/3

2. P(2 U 5 U 6) = 5/6

3. P(1 U 2) = 1/3

4. P(4) = 0

Step-by-step explanation:

As the number cube has 6 sides,

So,

Total outcomes = {1,2,5,5,6,6}

n(S) = 6

Now,

1. P(5U6) = P(5) + P(6)

P(5)=2/6=1/3

P(6)=2/6=1/3

P(5U6)= 1/3 + 1/3 = 2/3

2. P(2 U 5 U 6) = P(2) + P(5) + P(6)

P(2)=1/6

P(5)=2/6

P(6)=2/6

P(2 U 5 U 6) = 1/6 + 2/6 + 2/6

= 5/6

3. P(1 U 2) = P(1) + P(2)

P(1) = 1/6

P(2)= 1/6

P(1 U 2) = 1/6 + 1/6

= 2/6 = 1/3

4. P(4) = 0

As 4 is not in the outcomes, it's probability will be zero ..

5 0

3 years ago

Step 3: Now multiply the values to get 14,918,904,000. That is almost 15 billion passwords.

Troyanec [42]

The number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600

<h3>How to determine the number of passwords</h3>

The given parameters are:

Password length = 8

Letters = 6

Numbers = 2

There are 26 letters and 10 digits.

Since the letters and the numbers can be repeated, then the number of passwords is:

Count = 26^6 * 10^2

Evaluate

Count = 30891577600

Hence, the number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600