With what? I don't see a question
9514 1404 393
Answer:
Step-by-step explanation:
Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...
x + y = 85 . . . . . combined weight of a large and small box
70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes
We can subtract 50 times the first equation from the second to find the weight of a large box.
(70x +50y) -50(x +y) = (5350) -50(85)
20x = 1100 . . . . simplify
x = 55 . . . . . . . divide by 20
Using this in the first equation, we can find the weight of a small box.
55 +y = 85
y = 30 . . . . . . . subtract 55
A large box weighs 55 pounds; a small box weighs 30 pounds.
Answer:

option B is correct
Step-by-step explanation:
We have 5 spaces in the license plate:
_ _ _ _ _
we have 26 available letters, and 10 available numbers.
starting with letters:
- how many choices do i have to place the 1st letter? 26.
26 _ _ _ _
- how many choices do i have to place the 2nd letter? 26 (since we're allowed to repeat letters)
26 26 _ _ _
- how many choices do i have to place the 3rd letter? 26
26 26 26 _ _
we've used all the places for letters, (note: the exact position of the letters doesn't matter here, the first letter could've been placed anywhere in _ _ _ _ _, but the amount of possible choices for letters would always be 26).
let's move on to numbers.
- how many choices do i have to place the 1st number? 10
26 26 26 10 _
- how many choices do i have to place the 2nd number? 10
26 26 26 10 10
we've completed our number plate. Next we'll simply multiply all these numbers to get all the possible arrangements in which numbers and letters can be displayed on a license place.

option B is correct
Answer: 1/50 or 0.02
Step-by-step explanation:
30 / 1500 = 1/50
1/50 or 0.02
We know that the area of a circle in terms of π will be πr². However the area with respect to the diameter will be a different story. The first step here is to find a function relating the area and diameter of any circle --- ( 1 )
For any circle the diameter is 2 times the radius,
d = 2r
Therefore r = d / 2, which gives us the following formula through substitution.
A = π(d / 2)² = πd² / 4
<u>Hence the area of a circle as the function of it's diameter is A = πd² / 4. You can also say f(d) = πd² / 4.</u>
Now we can substitute " d " as 4, solving for the area ( A ) or f(4) --- ( 2 )
f(4) = π(4)² / 4 = 16π / 4 = 4π - <u>This makes the area of circle present with a diameter of 4 inches, 4π.</u>