The equation that matches the given situation is
. So, after two moves, Eric's elevation changed
meters above.
We know that in mathematics, subtraction means removing something from a group or a number of things. What is left in the group gets smaller when we subtract. The minuend is the first element we use. The subtrahend is the part that is being removed. The difference is the portion that remains after subtraction.
Assume that "negative" means climbing down and "positive" means climbing up. We must locate the elevation change in this area.
Given that Eric climbed straight down
meters. So we can write
.
Again Eric climbed straight up
meters. So, we can write
.
Then the change = ![-2\frac{1}{2}+6\frac{3}{4}](https://tex.z-dn.net/?f=-2%5Cfrac%7B1%7D%7B2%7D%2B6%5Cfrac%7B3%7D%7B4%7D)
=![-\frac{(2*2+1)}{2}+\frac{6*4+3}{4}](https://tex.z-dn.net/?f=-%5Cfrac%7B%282%2A2%2B1%29%7D%7B2%7D%2B%5Cfrac%7B6%2A4%2B3%7D%7B4%7D)
=![-\frac{5}{2} +\frac{27}{4}](https://tex.z-dn.net/?f=-%5Cfrac%7B5%7D%7B2%7D%20%2B%5Cfrac%7B27%7D%7B4%7D)
=![\frac{-(5*2)+27}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B-%285%2A2%29%2B27%7D%7B4%7D)
=![\frac{-10+27}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B-10%2B27%7D%7B4%7D)
=![\frac{17}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B17%7D%7B4%7D)
=![\frac{4*4+1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B4%2A4%2B1%7D%7B4%7D)
=![4\frac{1}{4}](https://tex.z-dn.net/?f=4%5Cfrac%7B1%7D%7B4%7D)
Therefore, the equation that matches the given situation is
. So, after two moves, Eric's elevation changed
meters above.
Learn more about subtraction here -
brainly.com/question/24116578
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Answer:
9
Step-by-step explanation:
Answer:
Step-by-step explanation:
f(x) = 4-x
g(x) = h
+k
g(f(x)) = 2
-16x+26
so put f(x) in g(x)
h
+k
h((4-x)(4-x) + k
h(
-8x+16)+k
if h = 2 , then
2
-16x+32 + k
and we want 26 instead of 32 so subtract 6 so K = (-6)
2
-16x+32 + (-6)
2
-16x+32 - 6
2
-16x+26
h=2
k=(-6)
67% of x = 67......turn ur percent to a decimal..." of " means multiply
0.67x = 67
x = 67 / 0.67
x = 100 <==
![\bf \begin{cases} f(x)=2-x^{12}\\ g(x)=x^2-9\\ g(x)\div f(x)=\frac{g(x)}{f(x)} \end{cases}\implies \cfrac{x^2-9}{2-x^{12}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3D2-x%5E%7B12%7D%5C%5C%0Ag%28x%29%3Dx%5E2-9%5C%5C%0Ag%28x%29%5Cdiv%20f%28x%29%3D%5Cfrac%7Bg%28x%29%7D%7Bf%28x%29%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7Bx%5E2-9%7D%7B2-x%5E%7B12%7D%7D)
now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to
undefined.
now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.
![\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\ -------------------------------\\\\ \cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}](https://tex.z-dn.net/?f=%5Cbf%202-x%5E%7B12%7D%3D0%5Cimplies%202%3Dx%5E%7B12%7D%5Cimplies%20%5Cpm%5Csqrt%5B12%5D%7B2%7D%3Dx%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2-9%7D%7B2-x%5E%7B12%7D%7D%5Cqquad%20%5Cboxed%7Bx%3D%5Cpm%20%5Csqrt%5B12%5D%7B2%7D%7D%5Cqquad%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%28%5Cpm%5Csqrt%5B12%5D%7B2%7D%29%5E%7B12%7D%7D%5Cimplies%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%5Cboxed%7B2%7D%7D%5Cimplies%20%5Cstackrel%7Bund%20efined%7D%7B%5Ccfrac%7Bx%5E2-9%7D%7B0%7D%7D)
so, the domain is all real numbers EXCEPT that one.