Three integers have a mean of 7, a median of 5 and a range of 14. find the three integers

Mathematics

1 answer:

inna [77]3 years ago

4 0

Answer:

1, 5, 15

Step-by-step explanation:

First, I knew 5 was one of them because, 5 was the median and there were only three numbers. Next you have to find out the first and last number. Then, I looked at the mean. To get the mean, you have to add all of the numbers up and divide that by how many numbers there are. Since the mean is 7 and there are 3 numbers, I knew that the sum of all three numbers was 21 because, 21 divided by 3 is 7. Then I saw that the range of all the numbers was 14. The range is the difference between the largest and smallest number so I knew that the largest number subtracted by the smallest number had to be 14. Then, I saw that 15-1=14, and 1+5+15=21. So the three integers are, 1, 5 and 15.

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Which values of x and y would make the following expression represent a real number? (4+5i)(x+yi)

algol [13]

Answer:

So x=4 and y=-5 would work.

Step-by-step explanation:

When you multiply complex conjugates, you get a real result.

Example:

$(a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2$

I replace $i^2$ with -1 since $i^2=-1$ .

$a^2+b^2$ is a real number (there is no imaginary part, no i).

So what is the conjugate of 4+5i?

4-5i

So x=4 and y=-5 would work.

A multiple of the complex conjugate would work as well.

$(a+bi)(ca-cbi)$

$c(a+bi)(a-bi)$

$c(a^2-b^2i^2)$

$c(a^2+b^2)$

This is still a real number; there is no imaginary part,no i.

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3 years ago

Suppose that (x, -8/17) is a point in quadrant IV lying on the unit circle.

AlladinOne [14]

Since the point lies on the unit circle, it satisfies

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Solving for $x$ gives two possible solutions:

$x^2=1-\left(\dfrac8{17}\right)^2$
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Given that $x$ is in quadrant IV, you know that its value must be positive, so

$x=\sqrt{1-\left(\dfrac8{17}\right)^2}=\sqrt{\dfrac{225}{289}}$