A local road has a grade of 10%. The grade of a road is its slope expressed as a percent. Find the slope of the road as fraction, and then simplify.

mangicele4s
2022-09-22
Answered

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Ruben Horn

Answered 2022-09-23
Author has **7** answers

Grade of road is slope in percentage

So slope of a x% grade $$=\frac{x}{100}\phantom{\rule{0ex}{0ex}}\text{Slope}=\frac{10}{100}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\text{Slope}=0.1$$

So slope of a x% grade $$=\frac{x}{100}\phantom{\rule{0ex}{0ex}}\text{Slope}=\frac{10}{100}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\text{Slope}=0.1$$

asked 2021-10-23

use the properties of logarithms, given that $\mathrm{ln}\left(2\right)\approx 0.6931\text{}\text{and}\text{}\mathrm{ln}\left(3\right)\approx 1.0986$ to approximate the logarithm. Use a calculator to confirm your approximations.

(a)$\mathrm{ln}\left(0.75\right)\approx$

(b)$\mathrm{ln}\left(288\right)\approx$

(c)$\mathrm{ln}\left(\sqrt{3}\left\{24\right\}\right)\approx$

(d)$\mathrm{ln}\left(\frac{1}{6}\right)\approx$

(a)

(b)

(c)

(d)

asked 2022-03-17

Can anyone show me how to simplify this fraction:

$\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6};.$

What can be factored out and so forth?

Thanks.

What can be factored out and so forth?

Thanks.

asked 2021-11-18

Write the equation of a line perpendicular to the line:

$y=-xy=-x$

that goes through the point (1, -2).

Write your answer in slope-intercept form, using simplified fractions for the slope and intercept

that goes through the point (1, -2).

Write your answer in slope-intercept form, using simplified fractions for the slope and intercept

asked 2022-06-12

How exactly is this happening?

I was studying Derivative and my book says if:

$y=\sqrt{(1+{x}^{2}{)}^{3}}$

Then its derivative is:

${y}^{\prime}=\frac{3(2x)(1+{x}^{2}{)}^{2}}{2\sqrt{(1+{x}^{2}{)}^{-1}}}=3x\sqrt{(1+{x}^{2})}$

I can't understand how the writer has changed the first derivative fraction into the second one. In other words, how did he simplify?

Note: I'm really really basic, so please explain in details.

Thank you.

I was studying Derivative and my book says if:

$y=\sqrt{(1+{x}^{2}{)}^{3}}$

Then its derivative is:

${y}^{\prime}=\frac{3(2x)(1+{x}^{2}{)}^{2}}{2\sqrt{(1+{x}^{2}{)}^{-1}}}=3x\sqrt{(1+{x}^{2})}$

I can't understand how the writer has changed the first derivative fraction into the second one. In other words, how did he simplify?

Note: I'm really really basic, so please explain in details.

Thank you.

asked 2020-11-23

asked 2022-07-05

another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le \frac{1}{2}$

Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that

$\begin{array}{}\text{(1)}& {\displaystyle \frac{1}{xy+z}}+{\displaystyle \frac{1}{yz+x}}+{\displaystyle \frac{1}{zx+y}}\le {\displaystyle \frac{1}{2}}\end{array}$

The theory basis of speculation Use the following classical results

$xyz=2+x+y+z\u27f9xy+yz+zx\ge 2(x+y+z)$

Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that

$\begin{array}{}\text{(1)}& {\displaystyle \frac{1}{xy+z}}+{\displaystyle \frac{1}{yz+x}}+{\displaystyle \frac{1}{zx+y}}\le {\displaystyle \frac{1}{2}}\end{array}$

The theory basis of speculation Use the following classical results

$xyz=2+x+y+z\u27f9xy+yz+zx\ge 2(x+y+z)$

asked 2022-09-11

Struggling to finish an exponential equation

I have this exponential equation I tried solving but I'm stuck at the end.

The problem is this:

$${4}^{x}-{3}^{x-\frac{1}{2}}={3}^{x+\frac{1}{2}}-{2}^{2x-1}$$

And my approach is this:

${2}^{2x}+{2}^{2x-1}={3}^{x+\frac{1}{2}}+{3}^{x-\frac{1}{2}}$

${2}^{2x}(1+{2}^{-1})={3}^{x}({3}^{\frac{1}{2}}+{3}^{-\frac{1}{2}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\frac{{\sqrt{3}}^{2}+1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}\ast \frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction

$\frac{{2}^{2x}}{{3}^{x}}=\frac{4\sqrt{3}}{3}\ast \frac{2}{3}$

${\left(\frac{{2}^{2}}{3}\right)}^{x}=\frac{8\sqrt{3}}{9}$

Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?

I have this exponential equation I tried solving but I'm stuck at the end.

The problem is this:

$${4}^{x}-{3}^{x-\frac{1}{2}}={3}^{x+\frac{1}{2}}-{2}^{2x-1}$$

And my approach is this:

${2}^{2x}+{2}^{2x-1}={3}^{x+\frac{1}{2}}+{3}^{x-\frac{1}{2}}$

${2}^{2x}(1+{2}^{-1})={3}^{x}({3}^{\frac{1}{2}}+{3}^{-\frac{1}{2}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\frac{{\sqrt{3}}^{2}+1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}\ast \frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction

$\frac{{2}^{2x}}{{3}^{x}}=\frac{4\sqrt{3}}{3}\ast \frac{2}{3}$

${\left(\frac{{2}^{2}}{3}\right)}^{x}=\frac{8\sqrt{3}}{9}$

Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?