Answer:
-6 + 3x = -9

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And 6 to both sides:

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3x = -3

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Divide by 3 on both sides:

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x = -1

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**Answer: x = -1**

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And 6 to both sides:

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3x = -3

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Divide by 3 on both sides:

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x = -1

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Help asap i will get yelled at =( this is going to give yoyu 42 points

Help find the circumstance and area

A die is rolled twice.What is the probability that the first roll was a 6 and the second roll was an odd number?

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.A(n) = 1 + (n – 1)(–5.7)1, –21.8, –56–5.7, –21.8, –51.31, –16.1, –50.30, –17.1, –51.3If someone answers this, i'd like if someone could help me to know how to solve it too, please and thank you!

A business student at Nowledge College must complete a total of 65 courses to graduate. The number of busi-ness courses must be greater than or equal to 23. The number of nonbusiness courses must be greater than or equal to 20. The average business course requires a textbook costing $60 and 120 hours of study. Non-business courses require a textbook costing $24 and 200 hours of study. The student has $3,000 to spend on books.

Help find the circumstance and area

A die is rolled twice.What is the probability that the first roll was a 6 and the second roll was an odd number?

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.A(n) = 1 + (n – 1)(–5.7)1, –21.8, –56–5.7, –21.8, –51.31, –16.1, –50.30, –17.1, –51.3If someone answers this, i'd like if someone could help me to know how to solve it too, please and thank you!

A business student at Nowledge College must complete a total of 65 courses to graduate. The number of busi-ness courses must be greater than or equal to 23. The number of nonbusiness courses must be greater than or equal to 20. The average business course requires a textbook costing $60 and 120 hours of study. Non-business courses require a textbook costing $24 and 200 hours of study. The student has $3,000 to spend on books.

A square has a side length of 2x+1. If the perimeter is 28, what is x?

bearing in mind that a square has **4 equal sides**, and the perimeter is the sum of all sides.

**Answer:**

x=-2 hope this helps! :)

**Step-by-step explanation:**

first you move the constant to the right

3x+1=-5

then you calculate

3x=-5-1

finally you divide both sides to get x alone

x=-2

**Answer:**

x= -2

**Step-by-step explanation:**

**Step 1**- Distribute into the parenthesis.

(2)3x+(2)1= -10

**Step 2**- Multiply

6x+2= -10

**Step 3**- Subtract 2 to both sides.

6x+2= -10

-2 -2

**Step 4**- Divide both sides by 6

__6x__= __-12__

6 6

x= -2

The **value **of **variables **are,

⇒ x = 13.85

⇒ y = 16

A triangle is a three sided polygon, which has three vertices and three angles which has the sum 180 degrees.

We have to given that;

**Base **of triangle = 8

Now, We get;

⇒ cos 60° = 8 / y

⇒ 1/2 = 8/y

⇒ y = 8 × 2

⇒ y = 16

And, By **Pythagoras theorem** we get;

⇒ x² + 8² = y²

⇒ x² + 8² = 16²

⇒ x² + 64 = 256

⇒ x² = 256 - 64

⇒ x² = 192

⇒ x = √192

⇒ x = 13.85

Thus, The **value **of **variables **are,

⇒ x = 13.85

⇒ y = 16

Learn more about the **triangle **visit;

#SPJ2

y = 8/cos60 = 16

x = sqr(16^2 + 8^2) = 8sqr(3) = 13.86 approx

x = sqr(16^2 + 8^2) = 8sqr(3) = 13.86 approx

**Answer:**

See the attached picture for detailed answer.

**Step-by-step explanation:**

See the attached picture for detailed answer.

The **probability **question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.

The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact **expression **for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.

For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial **distribution **and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.

For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate **approximation**. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.

#SPJ3

3−(−5)−(−2)

Plot the solution on the number line.

**Answer: Its 10, so plot it on 10 in the number line**

**Step-by-step explanation: Since the subtraction signs cancel out, they make a addition sign. Henceforth its " 3+5+2" which would mean the sum is 10.**

1.5 I think would be the correct answer not sure