All categories

4 years ago

Which statement is the correct interpretation of the inequality –2 > –8?

Mathematics

1 answer:

daser333 [38]4 years ago

3 0

The correct interpretation of the inequality –2 > –8 is On a number line, −2 is located to the right of –8.

You might be interested in

The length of a rectangle is 6 meters less than twice the width. The area of the rectangle is 140m². Find the dimensions of the

DIA [1.3K]

Equation 1 is 2W - 6 = L
equation 2 is 140= W * L

L=10
W=8

8 0

3 years ago

Read 2 more answers

What's 4/6 + 5/8 ? can u write it out so i can show my work?

frutty [35]

Answer:

2 and 3/8

Step-by-step explanation:

4/6 = 6/8 so 5/8 + 6/8 = 2 and 3/8

8 0

3 years ago

Read 2 more answers

Simplify using the Laws of Exponents. <br><br> (6a2b6)3

lina2011 [118]

(6a²b⁶)³
(6a²b⁶)(6a²b⁶)(6a²b⁶)
(6 · 6 · 6)(a² · a² · a²)(b⁶ · b⁶ · b⁶)
216a⁶b¹⁸

5 0

3 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

Event A: Spinning violet or yellow on the spinner shown.

Studentka2010 [4]

Wheres the rest of the question, or is that it?

7 0

3 years ago