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4 years ago

15

If gas cost $1.35 per gallon and Jose has $81 budgeted for gasoline, what os the maximum distance in miles he can travel if his

car averages 34 miles per gallon?

1 answer:

MA_775_DIABLO [31]4 years ago

3 0

So you times 1.35 by 60 and It gives you 81 so its 60 times 34 = 2040

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Express each of the as the sum of a whole number and a fraction .

Alla [95]

im not 100% sure if my answers are right but heres what i got:

b. 3 and 1/2

d. 2 and 1/3 (or 2 3/9)

3 0

4 years ago

Read 2 more answers

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago

Also find the volume of the figure below?

yan [13]

Volume = LxWxH
Use that formula and I'm sure u'll find ur answer

8 0

3 years ago

Is 2 over 1 is the same thing as 1 over 4

tamaranim1 [39]

No is not the same

2/1 = 2

1/4 = .25

7 0

3 years ago

Find the value of x.

agasfer [191]

Answer:

I think the answer is 14.

Step-by-step explanation:

I'm not 100% sure but I'm pretty sure you times 7 by 2.

7 0

3 years ago