All categories

3 years ago

Multiply (3x-8)(2x^2+4-9)

Mathematics

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

6 x^ 3 − 16 x^ 2 − 15 x + 40

Step-by-step explanation:

You might be interested in

What are the coordinates of B' if the origin is the center of dilation and the scale factor is 1/4?

Deffense [45]

Answer:

B' (1, 2 )

Step-by-step explanation:

Since the centre of dilatation is the origin then multiply the coordinates of B by the scale factor.

B (4, 8 ) → B' (4 × $\frac{1}{4}$ , 8 × $\frac{1}{4}$ ) → B' (1, 2 )

7 0

2 years ago

Read 2 more answers

Solve for x.... -4+13x=12x+4

Sav [38]

You put all x on left side and numercial on the right side

13x-12x = 4+4

x = 8

5 0

3 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

Write an equation of the line that passes through the point (5,-4) and has a slope of 2.

Romashka-Z-Leto [24]

Answer:

Point slope formula: (y + 4) = 2(x - 5)

Slope intercept formula: y = 2x - 14

Step-by-step explanation:

Point slope formula: (y + 4) = 2(x - 5)

Slope intercept formula: y = 2x + b, -4 = 10 + b, b = -14

y = 2x - 14

6 0

4 years ago

Evaluate log 16.7. 1.2227 -0.7773 1.0277 2.2227

spin [16.1K]

If you have a calculator with log and expo function capabilities, type in:

log 16.7 ENTER

and the calculator will return 1.2227. Try it.

3 0

3 years ago