Point E\text{'s}E’sE, start text, apostrophe, s, end text xxx- and yyy-coordinates have different signs (one is positive and the other is negative). Neither coordinate is 000.

Where could point EEE be located on the coordinate plane?

Step-by-step explanation:

If point E's x and y coordinates have different sign, there are two possibilities.

1. x is positive and y is negative, (x,-y)

2. x is negative and y is positive, (x,-y)

1. x is positive and y is negative, (x,-y)

When this occurs, point E will be in the lower right of the origin line as seen in Figure 1 attached.

2. x is negative and y is positive, (-x,y)

When this occurs, point E will be in the upper left of the origin line as seen in Figure 2 attached.

1

step-by-step explanation:

you are throwing one dart

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2