Answer:
C. Yes, the density equals that of pure silver (10.5)
Step-by-step explanation:
Get the b all alone on one side of the inequality. To do this, we can divide both sides by 5.
5b / 5 ≤ -15 / 5
b ≤ -3
So, that means b has to be less than or equal to -3.
In interval notation it would look like [-3, -∞)
In a set builder notation: {x | x ∈ R, x ≤ -3}
Which is read as the set of all x, such that, x is a member of the real numbers and x is less than or equal to -3.
Answer:
Answer: 84 meters squared. (i.e. 84m^2).
Step-by-step explanation:
Area of the path = total area of path and plot - plot area only
First, let’s find the area of the plot:
20m x 20m = 400m^2
The path is 1m wide and runs all the way around, so, it describes an ‘outer square’ of 22m x 22m (2 x 1m larger because there are 1m of the path on each side).
The area of this ‘outer square’ is equivalent to the ‘total area of path and plot’
22m x 22m = 484m^2
So:
Area of the path = total area of path and plot - plot area only
= 484 - 400 = 84m^2
ANSWER:
A and B aren't parallel lines, as the alternate angles aren't equivalent to each other.
For m angle 1 -
Vertically opposite angles are equivalent to each other.
m angle 1 is vertically opposite m angle 4.
Therefore:
m angle 4 = 100
m angle 1 = m angle 4
m angle 1 = 100
For m angle 6 -
Co-interior angles add up to 180°.
m angle 4 and m angle 6 are co-interior angles.
Therefore:
m angle 4 = 100
m angle 4 + m angle 6 = 180
100 + m angle 6 = 180
m angle 6 = 180 - 100
m angle 6 = 80
For m angle 7 -
Vertically opposite angles are equivalent to each other.
m angle 6 is vertically opposite m angle 7.
Therefore:
m angle 6 = 80
m angle 7 = m angle 6
m angle 7 = 80
For m angle 8 -
Corresponding angles are equivalent to each other.
m angle 8 and m angle 4 are corresponding angles.
Therefore:
m angle 4 = 100
m angle 8 = m angle 4
m angle 8 = 100
Hence, the angles are as follows:
m angle 1 = 100
m angle 6 = 80
m angle 7 = 80
m angle 8 = 100
Hope this helps! <3
The answer is B....................................................................................
