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3 years ago

Plz help me get the answer

Mathematics

2 answers:

anyanavicka [17]3 years ago

6 0

Let's isolate (1/2)x. To do this, subtract 2 from both sides of the equation, obtaining (1/2)x < -7.

Next, multiply both sides by 2 to eliminate the fraction (1/2):

x < -14

miv72 [106K]3 years ago

4 0

Answer:

Step-by-step explanation:

$\dfrac{1}{2}x+2$

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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 <em>y</em> 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 |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 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.

7 0

3 years ago

24 x 10 x 90 = 90 x 2,400 True or False?

xenn [34]

False, 24 x 10 x 90 = 21,600, 90 x 2,400 = 216,000

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3 years ago

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Subtract polynomials:

Strike441 [17]

The answer to the question

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3 years ago

A grid model with 100 squares. 73 squares are shaded. What percent of the model is shaded? What is the equivalent ratio? What is

viva [34]

Answer:

73%

Step-by-step explanation:

First, you'll divide the total squares by 100 to get how much is one percent. Then, you divide the shaded squares by the value of 1%. And you'll get how many percents are shaded.

For equivalent decimal, you divide the number by 100.

Then for equivalent decimal, you get the percent and 100.

$p =100/100=1\\73/1=73\\=0.73\\=\frac{73}{100}$

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3 years ago

PLEASE HELP!!

I am Lyosha [343]

36 + 18 = 54

(1) 18/54 = 1/3 = 33%

43 + 24 = 67

(2) 24/67 = 36%

I hope these are right

3 0

3 years ago

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