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2 years ago

Batman took a bicycle trip. In all, he traveled 6 hours at a rate of 28 mph. What was his total distance ?

Mathematics

2 answers:

Mice21 [21]2 years ago

7 0

Distance=speed x time
so it would be 28 x 6=168

stich3 [128]2 years ago

3 0

Answer:

just multiply

Step-by-step explanation:

and ull get

168

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Aleksandr [31]

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3 years ago

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The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

$y=0.00035x^{2}$

Compute the first order derivative of y as follows:

$y=0.00035x^{2}$

$\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]$

$=2\cdot 0.00035x\\\\=0.0007x$

Now, it is provided that |x | ≤ 605.

⇒ -605 ≤ x ≤ 605

Compute the arc length as follows:

$\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx$

$=\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\$

Now, let

$x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u$

$\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u$

$={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}$

Plug in the solved integrals in Arc Length and solve as follows:

$\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\$

$=1245.253707795227\\\\\approx 1245.25$

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

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3 years ago

6y-6y-12y=please help me

Ray Of Light [21]

Hello.

The answer is

-12y

Combine Like Terms:=6y+−6y+−12y=(6y+−6y+−12y)=−12y

Have a nice day

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I need to find the absolute mean of this?

Slav-nsk [51]

Give more detail please this makes zero sense

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3 years ago

PLEASE HELP! What is AD? A. 15 B. 12 C. 3 D. 20

ki77a [65]

Answer:

A. 15

Step-by-step explanation:

To solve this you need to compare the lengths given to you in the question statement.

Because the lines originate from a single point, they're like triangles. We can easily see a triangle AGF and a triangle ADE, right?

Both triangles are similar triangles, so we can see triangle ADE as a larger version of angle AGF.

They give you the dimension of A F and A E (through A F + F E) to establish a ratio... and they give you A G, asking for A D.

So, A F = 16, A E = 20 (16 + 4), A G = 12.

Since A D is to A G what A E is to A F, we can easily make the following cross-multiplication:

$\frac{A D}{A G} = \frac{A E}{A F}$

So, A D = (A G * A E)/A F

A D = (12 * 20) / 16 = 15

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3 years ago