Answer:
Step-by-step explanation:
The distance is always positive and is same as the absolute value of subtraction of the two numbers on the line.
<u>This can be shown as:</u>
- |-3/2 - 6| =
- |-1.5 - 6| =
- |-7.5| =
- 7.5
or
- | 6 - (-3/2)| =
- |6 + 1.5| =
- |7.5| =
- 7.5
Whats the graph ??
i need it to find the answer
Something that a right triangle is characterised by is the fact that we may use Pythagoras' theorem to find the length of any one of its sides, given that we know the length of the other two sides. Here, we know the length of the hypotenuse and one other side, therefor we can easily use the theorem to solve for the remaining side.
Now, Pythagoras' Theorem is defined as follows:
c^2 = a^2 + b^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.
Given that we know that c = 24 and a = 8, we can find b by substituting c and a into the formula we defined above:
c^2 = a^2 + b^2
24^2 = 8^2 + b^2 (Substitute c = 24 and a = 8)
b^2 = 24^2 - 8^2 (Subtract 8^2 from both sides)
b = √(24^2 - 8^2) (Take the square root of both sides)
b = √512 (Evaluate 24^2 - 8^2)
b = 16√2 (Simplify √512)
= 22.627 (to three decimal places)
I wasn't sure about whether by 'approximate length' you meant for the length to be rounded to a certain number of decimal places or whether you were meant to do more of an estimate based on your knowledge of surds and powers. If you need any more clarification however don't hesitate to comment below.
Answer:
1. Number line 2
2. Number line 1
3. Number line 4
4. Number line 3
Step-by-step explanation:
1. x – 99 ≤ -104
Solving by adding +99 on both sides
x - 99 +99 ≤ -104 +99
x ≤ -5
Number line 2 represent x ≤ -5
2. x – 51 ≤ -43
Adding +51 on both sides
x -51 +51 ≤ -43 +51
x ≤ 8
Number line 1 represent x ≤ 8
3. 150 + x ≤ 144
Adding -150 on both sides
150 + x -150 ≤ 144 -150
x ≤ -6
Number line 4 represent x ≤ -6
4. 75 < 69 – x
Adding +x on both sides
75 + x < 69 -x +x
x < 69 -75
x < -6
Number line 3 represent x < -6
What is the question that u need help with
