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umka2103 [35]
2 years ago
15

SOMEONE PLSSS HELP!!! Explain the difference in the graphs of Car #1 and Car #2. What is the Slope of each? Why is one line stee

per than the other? What does this mean in terms of the Car's Fuel Efficiency or miles per gallon?

Mathematics
1 answer:
german2 years ago
7 0

Answer:

Car number 2 has a slope of exactly 2 while Car number one has a slope of 2.3 because it is not exactly on the 2 point line. Car 2 is steeper and Car two has better fuel efficiency


Hope this helps! <3



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What is the range of a function y=|x+3| -3 if the domain is x€R
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Answer:

y ≥−3

Step-by-step explanation:

1 By inspecting the graph, the range is:

y\ge -3y≥−3

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Find all real solutions of the equation. (if there is no real solution, enter no real solution.) 35x2 12x − 36
solmaris [256]
Important:  35x2 12x − 36 is not an equation; to obtain an equation, you must set <span>35x2 12x − 36 = to 0:

</span><span>35x^2 + 12x − 36 = 0.  Also, "the square of x" must be represented by "x^2," and you must insert either the + or the - sign in front of the term 12x.

I am assuming that you meant   </span>35x^2 + 12x − 36 = 0.

If that is indeed the case, then a = 35, b = 12 and c = -36

and the discriminant is b^2 - 4ac, or (12)^2 - 4(35)(-36) = 5184.

Since the discriminant is positive, this equation has 2 real, unequal solutions.  They are
        -12 plus or minus sqrt(5184)
x = -----------------------------------------
                         2(35)
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If 84 is divisible by 2, then 84 is an even number
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High population density can cause increases in competition for resources, such as food and shelter. The table shows the number o
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Read 2 more answers
Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6
Anton [14]

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

V= \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder,x^{2}+y^{2} =1 and the ellipsoid, 4x^{2}+ 4y^{2} + z^{2} =64

In polar coordinates, we know that

x^{2}+y^{2} =r^{2}

So, the ellipsoid gives

4{(x^{2}+ y^{2)} + z^{2} =64

4(r^{2}) + z^{2} = 64

z^{2} = 64- 4(r^{2})

z=± \sqrt{64-4r^{2} }

So, the volume of the solid is given by:

V= \int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta

= 2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta

To solve the integral take, 64-4r^{2} = t

dt= -8rdr

rdr = \frac{-1}{8} dt

So, the integral  \int\ r\sqrt{64-4r^{2} } rdr become

=\int\ \sqrt{t } \frac{-1}{8} dt

= \frac{-1}{12} t^{3/2}

=\frac{-1}{12} (64-4r^{2}) ^{3/2}

so on applying the limit, the volume becomes

V= 2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta

=\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta

V = \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Since, further the integral isn't having any term of \theta.

we will end here.

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0
1 year ago
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