Answer:
right square pyramid A≈742.98
Step-by-step explanation:
Height ; 40
base area ; 70
First simplifying step by step.
<span>16x^5+12xy-9y^5
Answer is = </span><span><span><span>16<span>x^5</span></span>+<span><span>12x</span>y</span></span>+</span>−<span>9<span>y<span>5
It will help you.</span></span></span>
Answer:
Step-by-step explanation:
Substitute 4x for Y: 4x = -2x -6
Now add 2x to both sides
(due to reciprocal, the opposite of -2x is positive 2x): 4x + 2x = -2x +2x -6
Add common like terms
(4x + 2x = 6x and -2x + 2x cancel out leaving -6 alone): 6x = 6
Now get X alone by dividing the 6 to both sides: 6x/6 = -6/6
Since you're dividing a negative and a positive, the outcome will be a negative so the answer is: X= -6
Hope this helps :)
![\bf \begin{array}{llll} &[(-6,2),(2,3),(1,1),(-7,2),(4,2)]\\\\ inverse& [(2,-6),(3,2),(1,1),(2,-7),(2,4)] \end{array} \\\\\\ \textit{is the original a one-to-one?}\qquad \stackrel{rep eated~y-values}{(-6,\stackrel{\downarrow }{2}),(2,3),(1,1),(-7,\stackrel{\downarrow }{2}),(4,\stackrel{\downarrow }{2})}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0A%26%5B%28-6%2C2%29%2C%282%2C3%29%2C%281%2C1%29%2C%28-7%2C2%29%2C%284%2C2%29%5D%5C%5C%5C%5C%0Ainverse%26%20%5B%282%2C-6%29%2C%283%2C2%29%2C%281%2C1%29%2C%282%2C-7%29%2C%282%2C4%29%5D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bis%20the%20original%20a%20one-to-one%3F%7D%5Cqquad%20%5Cstackrel%7Brep%20eated~y-values%7D%7B%28-6%2C%5Cstackrel%7B%5Cdownarrow%20%7D%7B2%7D%29%2C%282%2C3%29%2C%281%2C1%29%2C%28-7%2C%5Cstackrel%7B%5Cdownarrow%20%7D%7B2%7D%29%2C%284%2C%5Cstackrel%7B%5Cdownarrow%20%7D%7B2%7D%29%7D)
notice, the inverse set is just, the same set with the x,y turned to y,x, backwards.
is it a one-to-one? well, for a set to be a one-to-one, it must not have any x-repeats, that is, the value of the first in the pairs must not repeat, and it also must not have any y-repeats, namely the value of the second in the pairs must not repeat.
I don’t know:( But wish I could help. You can come check out some of my questions too!
