Consider the point on the polar graph below. Represent the point's location in the polar form (r,θ) where 0≤r<5 and 0≤θ&l t;2π.
(r,θ)=

Represent the point's location in the polar form (r,θ) where 0≤r<5and 2π≤θ&l t;4π.
(r,θ)=

Represent the point's location in the polar form (r,θ) where 0≤r<5 and −2π≤θ<0.
(r,θ)=

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37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)