Which category does each card go into?

Mathematics

2 answers:

WINSTONCH [101]3 years ago

7 0

Step-by-step explanation:

E3, E6, E11, E7: no solution

E1, E12 : 1 solution

E2, E8, E10 : infinite solutions

ad-work [718]3 years ago

5 0

Answer:

See attached

Step-by-step explanation:

The answer is attached

Infinitely many solutions

E2, E8, E10

One solution

E1, E7, E12

No solution

E3, E6, E11

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The inequality 10.45b + 56.50 < 292.67 is used to find the number of boxes (b) that can be loaded on a truck without exceedin

PSYCHO15rus [73]

We solve the inequality by subtracting 56.50 from both sides of the equation,
10.45b + 56.50 - 56.50 < 292.67 - 56.50
10.45b < 236.17
Then, divide both sides of the inequality by 10.45
b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22.

6 0

3 years ago

Read 2 more answers

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

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