All categories

3 years ago

Help me please! Correct answer get's brainliest :D

Mathematics

1 answer:

asambeis [7]3 years ago

5 0

answer:

step by step explanation:

make all numbers the same decimal,fraction or percentage and then compare

3/8=37.5%

-2.8=20.8%

0.053=5.3%

-2.8%

ascending means from smallest to biggest so the answer would be 0.053

You might be interested in

Determine the input value for which the statement

Phoenix [80]

In this question , we have a graph given, and we have to find the x coordinate of the intersection point .

From the graph , the input value is approximately 3.3 .

In the graph,

$f(x) =3$

And for g(x), we need the slope and y intercept .

Slope is the ratio of rise and run .

Here rise equals 3 units and run equals 2 units. And the graph touch the y axis at -2 .

So the equation of g(x) is

$g(x) = \frac{3}{2}x -2$

We need to do

$f(x)= g(x)$

Substituting the values of the two functions, we will get

$3 = \frac{3}{2}x -2$

Adding 2 to both sides

$5 = \frac{3}{2}x$

Cross multiplication

$10 =3x
\\
x = \frac{10}{3}$

$x = 3.3$

So the input value is 3.3 approx

7 0

4 years ago

Read 2 more answers

Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π

Natasha2012 [34]

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$ . The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$