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makkiz [27]
3 years ago
9

How do you write a profit of $28

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0

Answer: Linear

Step-by-step explanation:

It wouldn't be exponential because there isn't exponent, and it won't be rational because there is no fraction, and it is not quadratic. Hope this helps :)

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15. The equation below defines z as a differentiable function of x and y. Find the value of dz/dy at the point (1, 1, 1).
MariettaO [177]

Isolate the term with z^2.

x^2 - 5y^2 + xyz^2 = y - 4 \implies xyz^2 = -x^2 + 5y^2 + y - 4

Differentiate both sides with respect to y.

\dfrac{\partial(xyz^2)}{\partial y} = \dfrac{\partial(-x^2 + 5y^2 + y - 4)}{\partial y}

By the product and chain rules,

xz^2 + 2xyz \dfrac{\partial z}{\partial y} = 10y + 1

Solve for the partial derivative, then evaluate at (x,y,z) = (1,1,1).

\dfrac{\partial z}{\partial y} = \dfrac{10y + 1 - xz^2}{2xyz}

\dfrac{\partial z}{\partial y} \bigg|_{x=1,y=1,z=1} = \dfrac{10 + 1 - 1}{2} = \boxed{5}

4 0
2 years ago
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 7x − 5x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
Margaret [11]

Answer:

Therefore the solution is = k{-7,5,1} where k ∈R

Step-by-step explanation:

Given that,

f₁(x) =x

f₂(x)= x²

f₃(x)= 7x - 5x²

Also,

g(x) = c₁f₁(x)+c₂f₂(x)+c₃f₃(x)

Putting the values of f₁(x), f₂(x) and f₃(x).

g(x) = c₁.x+c₂x²+c₃(7x-5x²)

Given condition that g(x)= 0

∴ c₁.x+c₂x²+c₃(7x-5x²)=0

⇒(c₁+7c₃)x +(c₂-5c₃)x² = 0

Comparing the coefficients of x and x²

∴c₁+7c₃=0               and       c₂-5c₃ =0

\Rightarrow c_1 =-7c_3                     \Rightarrow c_2=5c_3

Let c₃= k   [k∈R]  

Then c₁ = -7k   and   c₂=5k

Therefore the solution is = { c₁,c₂,c₃}  

                                           = {-7k, 5k, k}

                                            =k{-7,5,1}

5 0
3 years ago
If triangle fgh m f = m h, gf = x + 40, hf =3x - 20, and gh = 2x + 20. The length of GH is
Furkat [3]

Answer:

d

Step-by-step explanation:

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Answer:

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Step-by-step explanation

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Mrs. Walker is tracking the odometer readings on her car for one week to see how much she uses her car for work and how much for
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Step-by-step explanation:

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4 years ago
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