She has to buy both binders and notebooks. So, you have to take into account that she has to have both. The closest you can get to $20 while still getting notebooks, is to buy 4 binders. 4 times 4 equals 16. So, she can get 4 binders and 2 notebooks, because then, 2 times 2 equals 4 and 16 plus 4 equals 20.
18 would have to be the mean because of the average
Answer:
I think this is the right answer but last two signs are incorrect
Hope this will help
Greetings!To find the length of any side of a
right triangle, you can use the
Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
Input the information from the diagram into the formula:
Expand each term:



Combine like terms:
Add -169 to both sides:

Factor out the Common Term (2):
Factor the Complex Trinomial:



Set Factors to equal
0:


or


However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)
The Solution Is:

I hope this helped!
-Benjamin
Answer:
a) SPAZ is equilateral.
b) Diagonals SA and PZ are perpendicular to each other.
c) Diagonals SA and PZ bisect each other.
Step-by-step explanation:
At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.
a) If figure is equilateral, then SP = PA = AZ = ZS:
![SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}](https://tex.z-dn.net/?f=SP%20%3D%20%5Csqrt%7B%5B4-%28-4%29%5D%5E%7B2%7D%2B%5B%28-2%29-%28-4%29%5D%5E%7B2%7D%7D)

![PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}](https://tex.z-dn.net/?f=PA%20%3D%20%5Csqrt%7B%286-4%29%5E%7B2%7D%2B%5B6-%28-2%29%5D%5E%7B2%7D%7D)



![ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}](https://tex.z-dn.net/?f=ZS%20%3D%20%5Csqrt%7B%5B-4-%28-2%29%5D%5E%7B2%7D%2B%28-4-4%29%5E%7B2%7D%7D)

Therefore, SPAZ is equilateral.
b) We use the slope formula to determine the inclination of diagonals SA and PZ:




Since
, diagonals SA and PZ are perpendicular to each other.
c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:








Then, the diagonals SA and PZ bisect each other.