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Afina-wow [57]
3 years ago
6

What is two over four divided by one over two

Mathematics
2 answers:
Natasha2012 [34]3 years ago
5 0

Answer:

0.25

Step-by-step explanation:

Svet_ta [14]3 years ago
3 0

Answer:

0.25

Step-by-step explanation:

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1. Find a formula for the nth term of the arithmetic sequence:
Dahasolnce [82]

Answer:

Step-by-step explanation:

This is an arithmetic progression.

difference d = (-8) - (-2) = (-8) + 2 = (-6).

nth term = a + (n-1)d

6 0
3 years ago
Can someone please help me?
Nesterboy [21]

Answer:

2

Step-by-step explanation:

i dont know

8 0
3 years ago
I need help with this problem and a few others
Leokris [45]
The answer you chose is correct. Those three values are more than -6.
6 0
3 years ago
Last month in the village of 125 people we’re issued a parking ticket. This month only 83 people were issued one. What was the p
natita [175]

Answer:

the percent reduction is <u>33.6%</u> and the number of people who were issued a parking ticket are <u>42</u>.

Step-by-step explanation:

Given:

Last month in the village of 125 people were issued a parking ticket.

This month only 83 people were issued one.

Now, to find the percent reduction from this month to last month and the number of people who were issued a Parking ticket.

So, the reduction of the number of people who were issued a Parking ticket from this month to last month are:

125 people - 83 people = 42 people.

Thus, 42 people reduction from this month to last month.

Now, to get the percent reduction:

\frac{42}{125}\times 100.

=\frac{4200}{125}

=\ 33.6\%

Therefore, the percent reduction is 33.6% and the number of people who were issued a parking ticket are 42.

7 0
3 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
8 0
3 years ago
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