Answer:
1 a.) 2t – 2
1 b.) –9z + 4
2 a.) –2v + 3
2 b.) –13k – 9
3 a.) 7w − 2
3 b.) 10s
4 a.) 5
4 b.) –3p + 11
5 a.) –29x + 9
b.) 15v – 60x – 2
a.) 25n + 2
b.) –15n
a.) –71x + 4
b.) –15b
Step-by-step explanation:
- <em>NOTE: I am kind of confused about the way this question is formatted, so I tried to answer everything I could decipher.</em>
1 a.) –2t + 5t – 2 – t → 2t – 2
1 b.) –5z + 4 – 4z → –9z + 4
2 a.) 8v + 3 – 10v → –2v + 3
2 b.) –6 – 6k – 3 – 7k → –13k – 9
3 a.) –6 – 3w + 4 + 10w → 7w − 2
3 b.) 9s + s → 10s
4 a.) 6 + 8b – 1 – 8b → 5
4 b.) 1p + 6 + 5 – 4p → –3p + 11
5 a.) 4x – 3x + 9 – 6x 5 → –29x + 9
b.) 9v + 6v – 2 – 10x 6 → 15v – 60x – 2
a.) n + 2 – (–4)n 6 → 25n + 2
b.) 6n – 3n 7 → –15n
a.) –4x – 4x + 4 – 9x 7 → –71x + 4
b.) –9b + (–6)b → –15b
times more stars are there in universe compared to human eye can see
<h3><u>
Solution:</u></h3>
Given that, conservative estimate of the number of stars in the universe is 
The average human can see about 3,000 stars at night with only their eyes
To find: Number of times more stars are there in the universe, compared to the stars a human can see
Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see
Then from given statement,
<em><u>Substituting given values we get,</u></em>
Thus times more stars are there in universe compared to human eye can see
Answer:
19/24
Step-by-step explanation:
5/12 + 3/8
LCM: 24
(5×2 + 3×3)/24
(10 + 9)/24
19/24
In geometry, a hypotenuse is the longest side of a right-angled triangle, the side opposite the right angle. The length of the hypotenuse can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides.
<em>It is useful in different situations involving the need to find distance or the measure of an angle.</em>
Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Take a square root of sum of squares: c = √(a² + b²)
