-74.25 or -72 1/4 (if ur multiplying by -11, if u subtracting -11 then it would be -4 1/4
Answer:
Check the attached graph
Step-by-step explanation:
Given equation is
.
Now we need to graph the given equation
..
Clearly it is not linear as equation is not of the form
. So we need to find some points on given equation
. by plugging some random values of x like -5,-3,1,... etc.
for x=-5, we get:
![g\left(x\right)=\frac{10}{x}](https://tex.z-dn.net/?f=g%5Cleft%28x%5Cright%29%3D%5Cfrac%7B10%7D%7Bx%7D)
.
Hence first point is at (-5,-2). Similarly we can find more points and graph those points.
x can't be 0 as division by 0 is not defined so it will have vertical asymptote at x=0.
Since degree of numerator is less than degree of denominator so y=0 will be horizontal asymptote.
Now join those points by a curved line to get final graph.
Answer:B
Step-by-step explanation:
Answer:
it's 2! ;)
Step-by-step explanation:
hope you are having a spectacular day and thankss!! - jada
Answer:
E=
Joules
Step-by-step explanation:
1) Olá! Adicionando dados faltantes ao enunciado, como a fórmula dada:
e corrigindo a quantidade de Energia inicial:
Vamos seguir aplicando a fórmula, lembrando das propriedades de logaritmo.
![M=\frac{2}{3}log(\frac{E}{E_{0}})\\9=\frac{2}{3}log(\frac{E}{10^{4,5}})*3\\27=2log(\frac{E}{10^{4.5}})\\log(\frac{E}{10^{4.5}})^{2}=27\\(\frac{E^{2}}{10^{9}})=10^{27}\\E^{2}=10^{27}*10^{9}\Rightarrow E=\sqrt{10^{36}}\rightarrow E=10^{18} Joules](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B2%7D%7B3%7Dlog%28%5Cfrac%7BE%7D%7BE_%7B0%7D%7D%29%5C%5C9%3D%5Cfrac%7B2%7D%7B3%7Dlog%28%5Cfrac%7BE%7D%7B10%5E%7B4%2C5%7D%7D%29%2A3%5C%5C27%3D2log%28%5Cfrac%7BE%7D%7B10%5E%7B4.5%7D%7D%29%5C%5Clog%28%5Cfrac%7BE%7D%7B10%5E%7B4.5%7D%7D%29%5E%7B2%7D%3D27%5C%5C%28%5Cfrac%7BE%5E%7B2%7D%7D%7B10%5E%7B9%7D%7D%29%3D10%5E%7B27%7D%5C%5CE%5E%7B2%7D%3D10%5E%7B27%7D%2A10%5E%7B9%7D%5CRightarrow%20E%3D%5Csqrt%7B10%5E%7B36%7D%7D%5Crightarrow%20E%3D10%5E%7B18%7D%20Joules)