Answer:
1 - expressions that are equal in value even though they are written in different ways, 2 - a(b + c) = ab + ac, or a(b + c) = ab - ac, 3 - a number that divides evenly into another number; a number multiplied to get a product, 4 - a single term; multiple terms connected by an addition or subtraction sign, 5 - a mathematical statement that shows two expressions are equal using an equal sign, 6 - an expression that uses variables to state a rule, 7 - a relation in which for any given input value, there is only one output value.
What do you call a bus full of white people? A Twinkie
Answer:
$420.
Step-by-step explanation:
What is 30% off 1400 Dollars. The easiest way of calculating discount is, in this case, to multiply the normal price $1400 by 30 then divide it by one hundred. So, the discount is equal to $420.
Answer:
I'm sorry i think u forgot to add the image
Step-by-step explanation:
Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:
<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.
<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.
If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.
