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creativ13 [48]
3 years ago
9

Ibrahim wants to give each of his six friends 1 2/3 candy bars. How many candy bars does he need to buy?

Mathematics
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

10 candy bars

Step-by-step explanation:

Since each of his friends needs a candy bar, you need to multiply 6 and 1 ⅔ together.

First, convert 1 ⅔ into an improper fraction: this gives us ⁵⁄₃.

Next, multiply ⁶⁄₁ and ⁵⁄₃ together. To do this, you can visualize 6 as ⁶⁄₁ (which is the same thing). Now you have ⁶⁄₁ x ⁵⁄₃.

<u>Simplify:</u>

The 6 in the numerator and the 3 in the denominator cancel out. This gives us ²⁄₁ x ⁵⁄₁ , which is 2 x 5.

2 x 5 = 10

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<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%20%5Cinfty%20%20%5Cfrac%7B%20%5Csqrt%5B%20%20%5Cscriptsize%5Cphi%
Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx

Replace x \to x^{\frac1\phi} = x^{\phi-1} :

I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx

Split the integral at x = 1. For the integral over [1, ∞), substitute x \to \frac1x :

\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx

The integrals involving tan⁻¹ disappear, and we're left with

I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}

8 0
2 years ago
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
Will give brainliest if correct Pls help thanks :)
lidiya [134]
Domain:
{-1, 3, 6}

Range:
{4, 5, 6}
6 0
3 years ago
Solve the problem below
Sliva [168]

Answer:

T = 60 degrees

Step-by-step explanation:

The dotted line is the height so it is a right angle

We are able to use trig functions since this is a right triangle

cos T = adj side / hyp

cos T = a/b

cos T = 8 sqrt(2) / 16 sqrt(2)

cos T = 1/2

Taking the inverse of each side

cos^-1 ( cosT) = cos^-1 ( 1/2)

T = 60 degrees

3 0
3 years ago
Read 2 more answers
Which best describes the transformation that occurs from the graph of f(x) = x2 to g(x) = (x – 2)2 + 3?
Stels [109]
Right 2 up 3 because when the change happens within x the opposite result happens but for vertical it remains the same . So the answer is <span>right 2, up 3</span>
7 0
3 years ago
Read 2 more answers
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