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Natasha_Volkova [10]
3 years ago
11

T¹/³+t¹/²=12.Pleass solve for me​

Mathematics
2 answers:
mart [117]3 years ago
6 0
I think it’s 72/5 there’s the proof I hope im was helpful

ladessa [460]3 years ago
6 0

Answer:

t=64

Step-by-step explanation:

L.C.M. of 2 and 3=6

t^{\frac{1}{6} } =x\\squaring\\ t^{\frac{1}{3} } =x^{2} \\cubing\\t^{\frac{1}{2} } =x^{3} \\x^{2} +x^{3} =12\\when~ x=2\\2^2+2^3=4+8=12\\so~ x=2~ is~ one~ root.\\

By synthetic division.

2| 1   1    0   -12

  |     2    6   12

 -------------------

     1   3    6|0

x²+3x+6=0

disc.=b²-4ac=3²-4×1×6=9-24=-15<0

so roots are imaginary.

x=2

t^{\frac{1}{6} }  =2\\t=2^6=64

if the statement is as given by Sadievigil  then he or she is correct otherwise refer to my solution.

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

4 0
3 years ago
2 4/10 + 3 5/8 = what does it equal
photoshop1234 [79]
2 4/10 + 3 5/8
= 6 1/40
4 0
2 years ago
Read 2 more answers
How many 1/10 cubes are in 4 1/2
larisa [96]

Answer:

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Step-by-step explanation:

There are 10 cubes in each 1. so 10 x 4 = 40. 1/2 of 10 is 5 so 40+5 = 45.

4 0
3 years ago
I need help with this question
mr Goodwill [35]

Answer:

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Step-by-step explanation:

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3 years ago
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In-s [12.5K]

Answer:

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Step-by-step explanation:

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