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2 years ago

Which table represents a linear function?

Mathematics

2 answers:

ladessa [460]2 years ago

6 0

Answer:

(0,3), (1,1), (2,-1), (3,-3)

Step-by-step explanation:

good luck, i hope this helps :)

Kisachek [45]2 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

8 0

3 years ago

Please help asap will give brainliest if correct.

professor190 [17]

Answer:

x¹²,8x⁴,6x,2x,1

Step-by-step explanation:

x²×x⁶×x⁴=x¹²

x^4(8) =8x⁴

x^1(6) =6x

x^1(2) =2x

x^0 = 1

3 0

3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$