The coordinates of B would be (-1, -3)
<u>answer (in words)</u>
FALSE. the coordinate pair (5, 2) is not a solution to the equation
. in order to figure out whether or not the statement is true or false, plug the
and
values from the coordinate pair (5, 2) into the given equation,
. if both sides of the equation end up equal, the coordinate pair is a solution to the equation. if not, the coordinate pair is not a solution to that equation.
<em>(i hope i explained that well enough, i'm better at explaining it algebraically as opposed to putting it into words lol)</em>
<u>answer (algebraic/steps for solving)</u>
first, plug in 5 for
in the equation
.
⇒ 
then plug in 2 for
.
⇒ 
now your equation is
. all that's left to do is to simplify. you can do this in whatever order you'd like, but i'll start with multiplying 2 · 5.
⇒ 
multiply 3 · 2.
⇒ 
add 10 + 6.
⇒ 
16 and 10 are <em>not</em> equal, therefore (5, 2) is not a solution to the equation
. in order for a coordinate pair to be the solution to an equation, both sides of the equation need to end up equal after solving and simplifying.
i hope this helps! have a great rest of your day <3
Decreased as getting more for less/same amount makes it so the worth of euro has went down
<u>What we'll do:</u>
A method to find whether an object will float or sink in a liquid is to find it's specific gravity with respect to the liquid, The Specific Gravity tells us the percentage of object's mass which will be submerged in water
<u>Finding the Specific Gravity:</u>
<u>Density of Box:</u>
Density = Mass / Volume
Density = 80 grams / 78 cm³
Density = 1.02 grams/cm³
<u>Specific Gravity of Box with respect to water:</u>
Specific Gravity = Density of Box / Density of Water
we know that density of water is 1 gm/cm³, plugging the values
Specific Gravity = 1.02 / 1
Specific Gravity = 1.02
Hence, 102% of the object will be under water
Q. but I've only heard about 100% of something, how can something be more than 100%?
A. The answer is that the extra 2% over the 100% is providing to the object being even deeper underwater