(3.15x10^4) (5.15x10^-7 express the result as a decimal

loris [4]

Answer:

8

Step-by-step explanation:

If you go along the x-axis to 9 and go up till it meets the blue line and go left you will reach the integer 8 so f(9)=8

5 0

3 years ago

What is the area of the square

koban [17]

B. Each side of a square is equal. So every side is 5. Formula for area is length x width. So multiply 5 and 5. Answer is 25.

3 0

3 years ago

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Mrs Bean bought a business at the start of 2006

Luden [163]

$A = P(1+\frac{r}{n} )^{nt} \\P = 30000\\r = .02\\n = 1 time a year\\t = 2018 - 2006\\\\A = 30000(1 + .02/1)^{1*12} \\A = \$ 38047.25\\----------------\\Solve for 'r'\\95000 = 8000(1 + r/1)^{1*5} \\95000 = 8000(1+r)^{5} \\11.875 = (1 + r)^{5}\\1.64031 = 1+r\\$

I see I did something wrong in the 2nd part, just take the first question as the right one.

4 0

3 years ago

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Can two vertical angles form a linear pair?

weqwewe [10]

Adjacent Angles are two angles that share a common vertex, a common side, and no common interior points. (They share a vertex and side, but do not overlap.) A Linear Pair is two adjacentangles whose non-common sides form opposite rays. ... ∠1 and ∠3 are not vertical angles (they are a linear pair).

6 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