Please help!<br><br> select all that apply

Chef Rita Is cooking for a Sunday brunch.She knows that 22 pancakes can feed 8 people.She is wondering how many people she can f

AlexFokin [52]

Answer:

Chef Rita can feed 20 people with 55 pancakes.

Step-by-step explanation:

We can solve this problem using equivalent ratios. Since we know that Chef Rita can feed 8 people with 22 pancakes, we can set up the ratio:

$\frac{22pancakes}{8people}=\frac{22}{8}$

To solve how many people she would be able to feed with 55 pancakes, we can set ratios equal to each other and solve for 'x' which is the number of people.

$\frac{22}{8}=\frac{55}{x}$

We can see what factor we multiply 22 by to get 55 and then we multiply the denominator by the same factor to get our equivalent ratio. 55 ÷ 22 = 2.5. So, 8 x 2.5 = 20 people.

4 0

3 years ago

Read 2 more answers

The formula y y-y1=m(x-x1) please tell me what is m

grandymaker [24]

Answer:

There is no value in M

Step-by-step explanation:

The equation is undefined.

6 0

4 years ago

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

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