Answer:
SQUARE
Step-by-step explanation:
If Quadrilateral MNPQ has vertices M(4,0), N(0,6), P(-4,0) and Q(0, -6).
Find the following MN, NP, PQ and MQ
Using the formula for calculating the distance between two points
MN = √(6-0)²+(0-4)²
MN = √6²+4²
MN = √36+16
MN = √52
MN = 2√13
NP = √(0-6)²+(-4-0)²
NP = √6²+4²
NP = √36+16
NP = √52
NP = 2√13
PQ = √(-6-0)²+(0-(-4))²
PQ = √6²+4²
PQ = √36+16
PQ = √52
PQ = 2√13
MQ = √(-6-0)²+(0-4)²
MQ = √6²+4²
MQ = √36+16
MQ= √52
MQ = 2√13
Since the length of all the sides are equal, hence the shape is a SQUARE
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em>
Hello
DEF and IKL , equilateral =>
DE = 5x-12 = 3x-4
5x-3x= -4+12
2x= 8
x=4
DE = 5x-12 = 20-12 = 8
KL = 3x+24 = 10x+3
24 -3 = 10x-3x
21 = 7x
x= 21/7 = 3
KJ = 10x+3 = 30+3 = 33
Answer:
$111
Step-by-step explanation:
The bank balance is ...
$700 × (1 + 0.05) = $735
The cost of the computer is ...
$750 × (1 -0.20) × (1 +0.04) = $624
The remaining bank balance after paying for the computer is ...
$735 -624 = $111
_____
When you add a percentage, you effectively multiply by the sum of 1 and that percentage. The same is true if the amount "added" is negative (as for a discounted price).
(original amount) + (percentage)×(original amount)
Use the distributive property to factor out the original amount:
= (original amount)×(1 + percentage)
Answer:
Second choice:
Fifth choice:
Step-by-step explanation:
Let's look at choice 1.
I'm going to subtract 1 on both sides for the first equation giving me 
. I will replace the 
in the second equation with this substitution from equation 1.
Expand using the distributive property and the identity 
:
So this not the desired result.
Let's look at choice 2.
Solve the first equation for by dividing both sides by 2:
.
Let's plug this into equation 2:
This is the desired result.
Choice 3:
Solve the first equation for by adding 3 on both sides:
.
Plug into second equation:
Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:
Not the desired result.
Choice 4:
I'm going to solve the bottom equation for since I don't want to deal with square roots.
Add 3 on both sides:
Divide both sides by 2:
Plug into equation 1:
This is not the desired result because the 
variable will be squared now instead of the 
variable.
Choice 5:
Solve the first equation for by subtracting 1 on both sides:
.
Plug into equation 2:
Distribute and use the binomial square identity used earlier:
.
This is the desired result.
