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s2008m [1.1K]
3 years ago
13

1/9 (2m - 16) = 1/3 (2m + 4)

Mathematics
1 answer:
Ainat [17]3 years ago
4 0

Answer:

m=7\frac{5}{16}

Step-by-step explanation:

\frac{1}{9}(2m - 16)=\frac{1}{3}(2m + 4)

\frac{2}{9}m - 1\frac{7}{9} =\frac{2}{3}m+ 1\frac{1}{3}

\frac{2}{9}m - 1\frac{7}{9} -1\frac{1}{3} =\frac{2}{3}m+ 1\frac{1}{3}-1\frac{1}{3}

\frac{2}{9}m - 3\frac{1}{9}  =\frac{2}{3}m

\frac{2}{3}m-\frac{2}{9}m= \frac{2}{9}m -\frac{2}{9}m - 3\frac{1}{9}

\frac{4}{9}m= - 3\frac{1}{9}

\frac{4}{9}m/\frac{4}{9} = - 3\frac{1}{9}/\frac{4}{9}

m=7\frac{5}{16}

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Step-by-step explanation:

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3 years ago
If y =32 when x =8 find y when x =30
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Answer:

y = 120

Step-by-step explanation:

If y = 32 when x = 8, then <em>y</em> is 4 times the number of <em>x</em>. If x = 30, then (4)30 = 120.

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3 years ago
Find n for which the nth iteration by the Bisection Method guarantees to approximate the root of f(x) = 2x^2 − 3x − 2 on [−2, 1]
Lady_Fox [76]

Answer:

n = 29 iterations would be enough to obtain a root of f(x)=2x^2-3x-2 that is at most 10^{-8} away from the correct solution.

Step-by-step explanation:

You can use this formula which relates the number of iterations, n, required by the bisection method to converge to within an absolute error tolerance of ε starting from the initial interval (a, b).

n\geq \frac{log(\frac{b-a}{\epsilon} )}{log(2)}

We know

a = -2, b = 1 and ε = 10^{-8} so

n\geq \frac{log(\frac{1+2}{10^{-8}} )}{log(2)}\\n \geq 29

Thus, n = 29 iterations would be enough to obtain a root of f(x)=2x^2-3x-2 that is at most 10^{-8} away from the correct solution.

<u>You can prove this result by doing the computation as follows:</u>

From the information given we know:

  • f(x)=2x^2-3x-2
  • \epsilon = 10^{-8}

This is the algorithm for the Bisection method:

  1. Find two numbers <em>a</em> and <em>b</em> at which <em>f</em> has different signs.
  2. Define c=\frac{a+b}{2}
  3. If b-c\leq \epsilon then accept c as the root and stop
  4. If f(a)f(c)\leq 0 then set <em>c </em>as the new<em> b</em>. Otherwise, set <em>c </em>as the new <em>a</em>. Return to step 1.

We know that f(-2)=2(-2)^2-3(-2)-2=12 and f(1)=2(1)^2-3(1)-2=-3 so we take a=-2 and b=1 then c=\frac{-2+1}{2} =-0.5

Because 1-(-0.5)\geq 10^{-8} we set c=-0.5 as the new <em>b.</em>

The bisection algorithm is detailed in the following table.

After the 29 steps we have that 6\cdot 10^{-9}\leq 10^{-8} hence the required root approximation is c = -0.50

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