If I'm understanding your question, the drawing is 10cm and for every 10cm, the drawing is etched 1mm. This means that no matter how big the drawing gets, the actual object will always be bigger since it's 10cm larger always than the drawing. Hope this helped!
Answer:
D
Step-by-step explanation:
Here, we want to select which of the options explains the scenario in the question.
Firstly, 1,000 shares were purchased at $10 per share.
Mathematically the total amount of shares bought will be 10 * 1000 = $10,000
Also, we have the growth rate as 12.5% = 12.5/100 = 0.125
Thus, representing the scenario with a function, we have;
A(t) = 10,000 e0.125t
You just have to reverse the operations
1. -48-8= -56
2. -56/3= -18.66
ANSWER = -18.66
Answer:
6.13
Step-by-step explanation:
Using Sine Law we know that
Using your figure let's assign sides and angles:
A=? B = 60° C = 70°
a = 5 b = ? c = x
If we put that into our formula:
Notice that we have too many unknowns. We need to complete at least one ratio to do this, so how do we do this?
Notice we have 2 angles given, so we solve for the third angle. The sum of all angles in any triangle is always 180°
∠A + ∠B + ∠C= 180°
∠A + 60° + 70° = 180°
∠A + 130° = 180°
∠A = 180° - 130°
∠A = 50°
Now we can use this to solve for x.
So the closest answer would be 6.13
Problem 60) The only natural number is the number 62 (assuming the number is 62 and not some decimal value;it's impossible to tell because it appears to be cut off). Recall that the set of natural numbers is the set {1, 2, 3, 4, ...} which is synonymous with the set of counting numbers.
------------------------------------------------
Problem 62) Irrational numbers are numbers that we CANNOT write as fractions of whole numbers. In this case, pi is irrational and so is sqrt(17)
It is impossible to write pi as a ratio of two integers (in the form p/q where q is nonzero). The same applies to sqrt(17).
------------------------------------------------
Problem 64)
There are two nonnegative integers and they are: 0 and 62 (keeping with the same assumption made back in problem 60 above). All you do here is list whole numbers that aren't negative.
