The length of a rectangle is 14 feet. The width is 6 feet.

9u2v and which term are like terms? A.6u2, B.9uv2, C.15u2v, D.13u2v2

irina1246 [14]

It is essential that you use the symbol " ^ " to denote exponention. See the following changes I have made.

9u^2v and which term or terms are like terms? 15u^2v. Why? because both have "u^2" in them.

8 0

2 years ago

Read 2 more answers

8,36 : 1,6<br><br><br><br> pleaseeeeeeeeee

Lyrx [107]

Answer:

209 : 40 or 5.225 : 1

Step-by-step explanation:

Your calculator can tell you the ratio 8.36/1.60 is 5.225. Writing that decimal as a fraction, you can factor out 25 to get ...

8.36 : 1.6 = 5.225 : 1 = 5225 : 1000 = (25)(209) : (25)(40) = 209 : 40

4 0

2 years ago

a spinner has 1 yellow, 2 red, 3 blue and 2 green sections. What is the probability of not getting a yellow?

guajiro [1.7K]

Hello,

spinner has 1 yellow, 2 red, 3 blue and 2 green sections. What is the probability of not getting a yellow?

to find the probability to not get a yellow you have to know the total of color which is

1+2+3+2=8

and its 1 yellow

So the probability of not getting a yellow is 1/8

hope this help

4 0

3 years ago

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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

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