Which of these is not a variable term found in the binomial expansion of (a + b)8? (ignore coefficients.) a7b a3b5 ab6 b8

the thermometer is precise but not accurate

step-by-step explanation:

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

d.) yes, because the volume of the cone is greater than the volume of the sphere

step-by-step explanation: