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Pavlova-9 [17]
3 years ago
14

Lily's car has a fuel efficiency of 8 liters per 100 kilometers. What is the fuel efficiency of Lily's car in kilometers per lit

er? km L
Mathematics
2 answers:
AlekseyPX3 years ago
6 0

Answer:

The fuel efficiency of Lily's car in kilometers per liter is 12.5 km/litre.

Step-by-step explanation:

Ivahew [28]3 years ago
6 0
Lily’s car would be half a liter because the difference between kilometers and liters are far from apart
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Which of the following must describe an irrational number
zubka84 [21]

Answer:

An Irrational number is a number that cannot be expressed as the ratio of two integers.

Example: \pi

its numbers are infinite

6 0
3 years ago
Find the missing length
natka813 [3]

Answer:

x = 45

Step-by-step explanation:

Corresponding sides of similar triangles will be in same ratio.

\dfrac{x}{x+18}=\dfrac{20}{28}

Cross multiply

x*28 = 20*(x+18)   {Use distributive method}

28x = 20x + 360

28x - 20x = 360

8x = 360

 x = 360/8

x = 45

3 0
2 years ago
Over the course of 2 hours, Abe assembled 18 plastic toys. If Abe continues assembling toys at this rate, which proportion can b
Kaylis [27]

Answer: 126

Step-by-step explanation: multiply 7 times 18 toys 7 for the hours and 18 fr the toys and you will get 126

3 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
2x(x + 3) = 5x - 4standard form​
soldi70 [24.7K]

Answer:

did you make a typo? there is no real solution

Step-by-step explanation:

4 0
3 years ago
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