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Elodia [21]
3 years ago
15

line one passes through the points (-3,-4) and (3,8). line 2 passes through the pointe (5,-3) and (8,-3) find the slope of each

line. which line has the greater slope?
Mathematics
1 answer:
Oduvanchick [21]3 years ago
4 0

<u>The slope of line 1 is equal to 2.</u>

<u>The slope of line 2 is equal to 0</u>

<u>Line 1 has a larger slope, as the y value in line 2 does not change between x values.</u>

To find the slope of each line, we can use the slope formula.

\frac{y.2 - y.1}{x.2 - x.1} = m

Line 1:

8 - (-4) / 3 - (-3)

12 / 6

2

Line 2:

-3 - (-3) / 8 - 5

0 / 3

0

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The smiths went to Larosa's for dinner and spent $42.50 on dinner. There is a 6.5% tax and they want to give a tip of 20% before
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Answer:

Step-by-step explanation:

cost of dinner = $42.5

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Write the fraction or mixed number as a percent.<br><br> 2 3/10
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230%

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The perimeter of a shape will always be greater then the area of the shape
bazaltina [42]
Not necessarily. For example, you can have a rectangle whose dimensions are 12 and 6. Length is 12 and 6 is the width.

The area (length × width) of the rectangle will be 72.

The perimeter of the rectangle (2l + 2w) will be 36.

So, in this case, the area was double the perimeter and the perimeter was not greater than the area of the shape.

8 0
4 years ago
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
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