3 years ago

The one to one functions G and H are defined as follow

1 answer:

7 0

$G=\{(-8,1),(1,5),(4,-5),(8,-9)\}\\\\\Rightarrow\ \ \ G^{-1}=\{(1,-8),(5,1),(-5,4),(-9,8)\}\ \ \ \Rightarrow\ \ \ g^{-1}(1)=-8\\\\H(x)=3x+4\ \ \ and\ \ \ D_H=R\\\\ 3x=H(x)-4\ \ \ \Rightarrow\ \ \ x= \frac{H(x)-4}{3}\\\\h^{-1}(x)= \frac{x-4}{3} \ \ \ and\ \ \ D_{h^{-1}}=R\\\\\\(h^{-1}\circ h)(-2)=-2\\\\h^{-1}\bigg {(}h(-2)\bigg{)}=h^{-1}\bigg {(}3(-2)+4\bigg{)}=h^{-1}(-2)= \frac{-2-4}{3} = \frac{-6}{3} =-2$

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Solve 7b 3 - 4b = 3 - 3 (b 4)

frez [133]

7b + 3 - 4b = 3 - 3(b + 4)
3b + 3 = 3 - 3b - 12
3b + 3b = 3 - 12 - 3 = -12
6b = -12
b = -12/6 = -2
b = -2

8 0

3 years ago

Is the answer b? If not then what’s the answer

m_a_m_a [10]

I would say the answer is b, because it is a rhombus gape and all sides are congruent and there are 4 sides

3 0

4 years ago

Read 2 more answers

A radioactive substance decays exponentially. A scientist begins with 190 milligrams of a radioactive

worty [1.4K]

Answer: 65.1 (mg)

Step-by-step explanation:

$y=190*e^{-a*t}\\\\Calculate\ a:\\\\t=33,y=95(mg)\\\\190*e^{-a*33}=95\\\\e^{-a*33}=0.5\\\\-a*33=ln(0.5)\\\\a=\dfrac{ln(\dfrac{1}{2}) }{-33} \\\\a=0,02100... \approx{0.021}\\\\If\ t=51\ h, y=190*e^{-0.02100*51}=65.09167...\approx{65.1}$