Answer:
21 box of books can be brought up by the delivery person at one time
Step-by-step explanation:
Here, we want to know the number of box of books that the delivery person could bring up at one time.
Let the number of boxes be x , so the total mass of the boxes that could come up at a time will be x * 40 = 40x lb
Let’s add this to the mass of the delivery person = 150 lb
So the total mass going inside the lift would be 150 + 40x
So we have to equate this to the maximum capacity of the lift;
150 + 40x = 1020
40x = 1020 - 150
40x = 870
x = 870/40
x = 21.75
Now since we cannot have fractional boxes, the number of boxes that could come into the lift without exceeding the maximum capacity of the lift is 21
Answer:
x = √(28), or x = 5.292
Step-by-step explanation:
First you distribute the square to the values inside of the parentheses. so it ends up looking like this
5(x^2 - 25) - 9 = 6
add 9 to both sides
so its 5(x^2 - 25) = 15
Then distribute the multiplication of 5 to the contents within the parentheses
so it would be 5x^2 - 125 = 15
add 125 to both sides
you get 5x^2 = 140
divide by 5 on both sides
you get x^2=28
then, take the square root of both sides to reverse the square
√(x^2)=√(28)
and in the end you get x=5.292
but √(28) will probably be fine if your teacher doesn't want u to solve for that kind of stuff.
1=81 centimetres
1=82 centimetres
2=83 centimetres
3=84 centimetres
1=85 centimetres
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
