A^(2)+b^(2)+c^(4)=2020

a=√−c4−b2+2020 or a=−√−c4−b2+2020

b=√−c4−a2+2020 or b=−√−c4−a2+2020

c=cant find it

Step-by-step explanation:

Step 1 for a: Add -c^4 to both sides.

c4+a2+b2+−c4=2020+−c4

a2+b2=−c4+2020

Step 2: Add -b^2 to both sides.

a2+b2+−b2=−c4+2020+−b2

a2=−c4−b2+2020

Step 3: Take square root.

a=√−c4−b2+2020 or a=−√−c4−b2+2020

Step 1 for b: Add -c^4 to both sides.

c4+a2+b2+−c4=2020+−c4

a2+b2=−c4+2020

Step 2: Add -a^2 to both sides.

a2+b2+−a2=−c4+2020+−a2

b2=−c4−a2+2020

Step 3: Take square root.

b=√−c4−a2+2020 or b=−√−c4−a2+2020

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

days of the week

step-by-step explanation:

javier earned more money then kyle, so that won't always be a constant term, but the days of the week do not change so that term is always constant.

edit: i meant