Question 12 of 20 : Select the best answer for the question. 12. Which one of the following is an example of a repeating decimal? A. B. C. D.
Answers
Related Questions
Elliott's bed is 75 inches long. During his growth spurt, Elliott grew 6 inches. Before his growth spurt, he was 5 feet 11 inches tall. How much shorter is Elliott's bed than Elliott?inches
Y=x^3+125 Needs to be factored
F(x)=(lnx)²/2xgive f '(x)=??
Are all horizontal lines parallel
Answers
Cross multiply and get 1470/3
Your final answer is 490 deer.
Answers
Answer:
Aaron;s rate for mowing lawns is 0.5 acres per hour."as1.5 acres / 3 hours = 0.5 acres an hour2.5 acres / 5 hours = 0.5 acres an hour Option (B)
Step-by-step explanation:Aaron;s rate for mowing lawns is 0.5 acres per hour."as1.5 acres / 3 hours = 0.5 acres an hour2.5 acres / 5 hours = 0.5 acres an hour
Answers
An event that cannot possibly happen has a probability of zero. (examples below)
--If it is Thursday, the probability that tomorrow is Friday is certain, therefore the probability is 1.
--If you are sixteen, the probability of you turning seventeen on your next birthday is 1. It's a certain event.
Answers
Answer:
-11, -9, -7
the smallest: 2x+1 = -11
Step-by-step explanation:
{z - some integer}
2z+1 - the smallest number
7(2z+1) - seven times the smallest number
2z+1+2=2z+3 - the middle number
2z+3+2 = 2z+5 - the largest number
2(2z+5) - twice the largest number
7(2z+1) + 2(2z+5) - the sum of 7 times the smallest and twice the largest
7(2z+1) + 2(2z+5) = -91
14z + 7 + 4z + 10 = -91
-17 -17
18z = -108
÷18 ÷18
z = -6
2z+1 = 2(-6)+1 = -12 + 1 = -11
2z+3 = 2(-6)+3 = -12 + 3 = -9
2z+5 = 2(-6)+5 = -12 + 5 = -7
Final answer:
To solve this problem, we denote the smallest odd integer as 'x' and set up the equation 7x + 2(x+4) = -91. By solving this equation, we find that the smallest integer is -15.
Explanation:
To find the three consecutive odd integers, let us denote the smallest odd integer as x; therefore, the next two consecutive odd integers would be x + 2 and x + 4, respectively. The problem states that the sum of seven times the smallest integer and twice the largest integer equals to -91. So, we can translate this into the equation, 7x + 2(x+4) = -91. Solving this gives us x = -15. Hence the smallest integer is -15. The full solution is as follows:
- Let the smallest integer be x.
- Then the other two consecutive odd numbers are x + 2 and x + 4.
- According to the question, 7x + 2(x + 4) = -91.
- Solving for x, we get x = -15.
- Hence, the smallest odd integer sought is -15.
Learn more about Consecutive Odd Integers here:
#SPJ3
b. {2,2,4}
c. (1,2, sqrt of 3 wouldn't it be B because it has all even numbers?
Answers
The only set of numbers that could be the sides of a right triangle is **(b) {2, 2, 4}**.
The Pythagorean Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore, in order for a set of numbers to be the sides of a right triangle, the following equation must hold:
hypotenuse^2 = leg1^2 + leg2^2
Let's check each of the given sets:
(a) {2, 3, √13}
hypotenuse^2 = √13^2 = 13
leg1^2 = 2^2 = 4
leg2^2 = 3^2 = 9
13 ≠ 4 + 9
Therefore, {2, 3, √13} cannot be the sides of a right triangle.
(b) {2, 2, 4}
hypotenuse^2 = 4^2 = 16
leg1^2 = 2^2 = 4
leg2^2 = 2^2 = 4
16 = 4 + 4
Therefore, {2, 2, 4} can be the sides of a right triangle.
(c) {1, 2, √3}
hypotenuse^2 = √3^2 = 3
leg1^2 = 1^2 = 1
leg2^2 = 2^2 = 4
3 ≠ 1 + 4
Therefore, {1, 2, √3} cannot be the sides of a right triangle.
Therefore, the only set of numbers that could be the sides of a right triangle is **(b) {2, 2, 4}**.
Learn more about sides of a right triangle here:
#SPJ6
Final answer:
The set of numbers that could be the sides of a right triangle is {2,3, sqrt of 13}.
Explanation:
To determine whether a set of numbers could be the sides of a right triangle, we can use the Pythagorean theorem, which states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Let's check each set of numbers:
a. {2,3,√13}
b. {2,2,4}
c. (1,2,√3)
For set a, the sum of the squares of 2 and 3 is 13, which is equal to the square of √13. Therefore, set a could be the sides of a right triangle.
For set b, the sum of the squares of 2 and 2 is 8, which is not equal to the square of 4. Therefore, set b could not be the sides of a right triangle.
For set c, the sum of the squares of 1 and 2 is 5, which is not equal to the square of √3. Therefore, set c could not be the sides of a right triangle.
Therefore, the set of numbers that could be the sides of a right triangle is a. {2,3,√13}.
Learn more about right triangle here:
#SPJ12