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larisa86 [58]
3 years ago
8

Help me with this I will give brainliest

Mathematics
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

both linear equations will have the form

y = mx + c ( m is the slope and c the y- intercept )

Chef Mate

c(n) = mn + c

calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (2, 14) and (x₂, y₂ ) = (3, 18.5) ← 2 ordered pairs from the table

m = \frac{18.5-14}{3-2} = \frac{4.5}{1} = 4.5 , then

c(n) = 4.5n + c ( n represents ounces )

To find c substitute any ordered pair from the table into c(n)

using (4, 23 )

23 = 18 + c ⇒ c = 23 - 18 = 5

c(n) = 4.5n + 5 ← equation for Chef Mate

--------------------------------------------------------------------

Similarly for Grocery Gourmet

g(n) = mn + c

using (2, 23 ) and (3, 27 ) to find m

m = \frac{27-23}{3-2} = \frac{4}{1} = 4 , then

g(n) = 4n + c

using (4, 31 ) to find c

31 = 16 + c ⇒ c = 31 - 16 = 15

g(n) = 4n + 15 ← equation for Grocery Gourmet

------------------------------------------------------------------

Equate the 2 equations to find when cost is the same

4.5n + 5 = 4n + 15 ( subtract 4n from both sides )

0.5n + 5 = 15 ( subtract 5 from both sides )

0.5n = 10 ( divide both sides by 0.5 )

n = 20

Then when n = 20

c(n) = 4.5(20) + 5 = 90 + 5 = 95

g(n) = 4(20) + 15 = 80 + 15 = 95

Both Chef Mate and Grocery Gourmet charge $95 for 20 ounces of extract

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  (a) at t=2, the particle is moving toward the origin

  (b) a(t) = 70(8 -9t^2)/(3t^2 +8)^3; a(2) = -0.245

  (c) the particle approaches x = 1/3 as t gets large

Step-by-step explanation:

(a) The function x(t) is negative for -3 < t < 3, so at t = 2, the particle is to the left of the origin.

The velocity of the particle is given by the derivative of the position function:

  x'(t)=\dfrac{(3t^2+8)(2t)-(t^2-9)(6t)}{(3t^2+8)^2}=\dfrac{70t}{(3t^2+8)^2}

Then, at t=2, the expression is positive, indicating the particle is moving toward the origin.

__

(b) The acceleration is the derivative of the velocity, so is ...

  a(t)=\dfrac{70(3t^2+8)^2-(70t)(12t)(3t^2+8)}{(3t^2+8)^4}=\dfrac{70(3t^2+8-12t^2)}{(3t^2+8)^3}\\\\\boxed{a(t)=\dfrac{70(8-9t^2)}{(3t^2+8)^3}}

Then at t=2, the acceleration is ...

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  a(2) = -0.245

__

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