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3 years ago

Evaluate the expression 9P5

Mathematics

1 answer:

insens350 [35]3 years ago

7 0

Answer:

45p

Step-by-step explanation:

9p(5)

9p*5=45p

Please let me know if this wasn't how your problem was suppose to look

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Over a 24-hour period, the temperature in a town can be modeled by one period of a sinusoidal function. The temperature measures

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3 years ago

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The probability of drawing a purple from a bag is 3/8. What is the probability of the complement?

In-s [12.5K]

Answer: 5/8

Step-by-step explanation:

3/8 + 5/8

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3 years ago

Is the following statement correct? Explain.

Maksim231197 [3]

Answer: The answer is NO.

Step-by-step explanation: The given statement is -

If the graph of two equations are coincident lines, then that system of equations will have no solution.

We are to check whether the above statement is correct or not.

Any two equations having graphs as coincident lines are of the form -

$ax+by=c,\\\\dax+dby=dc,\\\\\textup{where}~~d\neq 1.$

If we take d = 1, then both the equations will be same.

Now, subtracting the second equation from first, we have

$a(1-d)x+b(1-d)y=c(1-d)\\\\\Rightarrow ax+by=c,~\textup{since}~d\neq 1,~\textup{so}~1-d\neq 0.$

Again, we will get the first equation, which is linear in two unknown variables. So, the system will have infinite number of solutions, which consists of the points lying on the line.

For example, see the attached figure, the graphs of following two equations is drawn and they are coincident. Also, the result is again the same straight line which has infinite number of points on it. These points makes the solution for the following system.

$2x+5y=10,\\\\6x+15y=30.$

Thus, the given statement is not correct.

4 0

3 years ago

The sum of two numbers is 144 and the difference is 81. what ar the two numbers

kykrilka [37]

The two numbers are 112.5 and 31.5.

5 0

3 years ago

Easy math, but not sure what I did wrong.

Ierofanga [76]

Answer:

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year. So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3. However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6. Same problem as before.

If a₁ = 3, then a₅ = 9. This is a possible solution.

If a₁ = 4, then a₅ = 12. If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year. So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0

3 years ago