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hoa [83]
3 years ago
15

What is the least common denominator of 7/10 and 4/9

Mathematics
1 answer:
melisa1 [442]3 years ago
6 0

Answer:

7/10 = 63/90

4/9 = 40/90

Step-by-step explanation:

Rewriting input as fractions if necessary:

7/10, 4/9

For the denominators (10, 9) the least common multiple (LCM) is 90.

LCM(10, 9)

Therefore, the least common denominator (LCD) is 90.

Calculations to rewrite the original inputs as equivalent fractions with the LCD:

7/10 = 7/10 × 9/9 = 63/90

4/9 = 4/9 × 10/10 = 40/90

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What is the value of 'x' in the equation shown below?<br> 2 (x + 8) - 4x = 10 x + 4
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Step-by-step explanation:

2x+16-4x=10x+4

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Step-by-step explanation:

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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
Solve this using a number line.<br> 3 1/8 ÷ 1/8
sp2606 [1]
Hey there!!

25 is the answer

Hope this helps. c:
4 0
3 years ago
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