MATHEMATICS HIGH SCHOOL

A triangle has side lengths of (1.3k+3.5m)(1.3k+3.5m) centimeters, (4.1k-1.6n)(4.1k−1.6n) centimeters, and (9.7n+4.4m)(9.7n+4.4m) centimeters. Which expression represents the perimeter, in centimeters, of the triangle?

Answers

Answer 1

Answer:

Answer:

(5.4k+7.9m+8.1n) centimeters

Step-by-step explanation:

Given the side length of a triangle;

S1 = (1.3k+3.5m) cm

S2 = (4.1k-1.6n) cm

S3 = (9.7n+4.4m) cm

Perimeter of the triangle = S1+S2 + S3

Perimeter of the triangle = (1.3k+3.5m) + (4.1k-1.6n) + (9.7n+4.4m)

Collect the like terms;

Perimeter of the triangle = 1.3k+4.1k+3.5m+4.4m-1.6n+9.7n

Perimeter of the triangle = 5.4k+7.9m+8.1n

Hence the expression that represents the perimeter of the triangle is (5.4k+7.9m+8.1n) centimeters

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1. Each of 9 friends chooses her favorite positive integera. The median of the chosen number is 91, what is the smallest the average of the 9 chosen numbers could be?

b. The median of the chosen number is 91, is there an limit to how large the aerge of the chosen numbers can be? If so, what is the largest the average can be?

c. The average of the chosen number is 91, what is the smallest the median of the 9 chosen numbers could be?

d. The average of the chosen numbers is 91. What is the largest the median of the chosen numbers could be?

Answers

Answer:

a) 1

b) There is no limit to which the largest number can be because we are only given information about the median.

c) 1

d) 90

Final answer:

The smallest average is 49 and the largest average is 91. The smallest median is 91 and the largest median is also 91.

Explanation:

a. Since the median is 91, at least 5 friends must choose numbers greater than or equal to 91, and at most 4 friends can choose numbers less than 91. To minimize the average, let's assume the four friends choose the smallest possible numbers less than 91 (1, 2, 3, and 4). The remaining five friends can then choose 91, 91, 91, 91, and 91. The average of the nine chosen numbers is (1 + 2 + 3 + 4 + 91 + 91 + 91 + 91 + 91)/9 = 49.

b. There is no limit to how large the average of the chosen numbers can be. The nine friends can all choose the same number, such as 91, which would make the average 91.

c. Since the average is 91, let's assume the eight friends choose the smallest possible numbers less than 91 (1, 2, 3, ..., 8). The remaining friend can then choose a number greater than or equal to 91. To minimize the median, the friend can choose the smallest possible number greater than or equal to 91, which is 91. So, the smallest median would be 91.

d. Since the average is 91, let's assume the eight friends choose the largest possible numbers less than 91 (84, 85, ..., 91). The remaining friend can then choose a number greater than or equal to 91. To maximize the median, the friend can choose the largest possible number greater than or equal to 91, which is 91. So, the largest median would also be 91.

Learn more about median and average of chosen numbers here:

brainly.com/question/33899535

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BEST ANSWER GETS A METAL!!!!Which pair of complex numbers has a real-number product?

(1 + 3i)(6i)
(1 + 3i)(2 – 3i)
(1 + 3i)(1 – 3i)
(1 + 3i)(3i)

Answers

we will proceed to verify each case to determine the solution

remember that

$i^(2) =-1$

case A) $(1 + 3i)(6i)$

applying distributive property

$(1 + 3i)(6i)=1*6i+3i*6i \n=6i+18i^(2)\n=6i+18*(-1)\n=6i-18$

$-18+6i$ ------> is not a real number

therefore

the case A) is not a real number product

case B) $(1 + 3i)(2-3i)$

applying distributive property

$(1 + 3i)(2-3i)=1*2+1*(-3i)+3i*(2)+3i*(-3i)\n=2-3i+6i-9i^(2)\n=2+3i-9*(-1) \n=11+3i$

$11+3i$ ------> is not a real number

therefore

the case B) is not a real number product

case C) $(1 + 3i)(1-3i)$

applying difference of square

$(1 + 3i)(1-3i)=(1)^(2)-(3i)^(2)\n=1-9i^(2)\n=1-9*(-1) \n=10$

$10$ ------> is a real number

therefore

the case C) is a real number product

case D) $(1 + 3i)(3i)$

applying distributive property

$(1 + 3i)(3i)=1*3i+3i*3i \n=3i+9i^(2)\n=3i+9*(-1) \n=3i-9$

$-9+3i$ ------> is not a real number

therefore

the case D) is not a real number product

the answer is

$(1 + 3i)(1-3i)$

The answer is (1 + 3i)(1 – 3i).

I'm having a hard time with fractions, what is the sum of 2/5 and 2/4

Answers

We will go nice and slow...

to add fractions, you have to have a common denominator.....

that being said....a common denominator of $(2)/(5)$ and $(2)/(4)$ is 20...

So change the fractions...

$(2)/(5) * (4)/(4) = (8)/(20)$
and
$(2)/(4) * (5)/(5) = (10)/(20)$

Now, you can add..

$(8)/(20)+ (10)/(20) = (18)/(20)$ ( $(9)/(10)$ simplified)

Thus, your answer.

well first you need to get the same bottom number so the bottom number is 20 and the top will be first fraction top is 8 other fraction is 10 so your answer is 18/20

What Is The Square Root of 9