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2 years ago

The most important component when reporting a numerical scientific result is the ______.

Mathematics

1 answer:

elena55 [62]2 years ago

4 0

The most important component when reporting a numerical scientific result is the unit.

This is further explained below.

<h3>What is a numerical scientific result?</h3>

Generally, Our work in numerical and scientific computing includes the creation, analysis, and implementation of computer algorithms with the purpose of finding mathematical solutions to issues originating in engineering and the sciences.

In conclusion, When presenting a numerical result from scientific research, the unit is the single most crucial component.

Read more about the numerical scientific result

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In the graphic below, the measure of angle ABC is:

ArbitrLikvidat [17]

(Angle ABD) + (Angle ABC) = 180
Angle ABC = 180 - (Angle ABD)
Angle ABC = 180 - 37
Angle ABC = 143

Answer: Choice A) 143 degrees

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3 years ago

A sphere with radius r has a volume of 100 square units. What is the volume of a sphere with a radius 3r ?

pshichka [43]

Answer:

300

Step-by-step explanation:

V=4/3 * pi* r^3

If you sub in 100 for V, do some manipulation you get r cubed to equal 23.78.

Now put the value of 23.78 in for r cubed and V comes out to be 300

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3 years ago

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What is twelve more than the quotient of a number and eight

ruslelena [56]

(x÷8)+12 hope this helps

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3 years ago

5+ 3(5)÷-2<br>order of operations<br>pemdas

lakkis [162]

Answer:

-2.5

Step-by-step explanation:

5+3(5)/-2

5+15/-2

5+-7.5=-2.5

3 0

3 years ago

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The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest t

Soloha48 [4]

The distance between a point $(x,y,z)$ and the origin is $\sqrt{x^2+y^2+z^2}$ . But since $(\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}$ , both $f(x)$ and $\sqrt{f(x)}$ have the same critical points, so we can consider instead the squared distance, $x^2+y^2+z^2$ .

We're looking for the extrema of $x^2+y^2+z^2$ subject to $x+y+2z=12$ and $z=x^2+y^2$ . The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)$

with critical points where the partial derivatives vanish:

$L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)$

$L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)$

$L_z=2z+2\lambda+\mu=0$

$L_\lambda=x+y+2z-12=0$

$L_\mu=z-x^2-y^2=0$

From the first two equations, it follows that $x=y$ .

Then in the last two equations,

$x+y+2z-12=0\implies x+z=6$

$z-x^2-y^2=0\implies z=2x^2$

$\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2$

If $x=\frac32$ , then $z=6-\frac32=\frac92$ ; if $x=-2$ , then $z=8$ .

So there are two critical points, $\left(\frac32,\frac32,\frac92\right)$ and $(-2,-2,8)$ .

Let $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ . We have a minimum distance of $f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2}$ and maximum distance of $f(-2,-2,8)=\boxed{6\sqrt2}$ .

8 0

4 years ago