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RideAnS [48]
1 year ago
15

Find 5/16 of £180 please help as soon as possible

Mathematics
1 answer:
Zanzabum1 year ago
3 0

Answer:

56.25

Step-by-step explanation:

so basically you divide 180 by the denominator of the fraction which is 16 and multiply by the numerator, so 180 divided by 16 times 5 cool i guess

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

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

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

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

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

8 0
3 years ago
How many solutions does x²+3x=3 have?
aev [14]

Answer:

2 (real) solutions.

Step-by-step explanation:

A quadratic always has two solutions, whether they are real or complex.

Sometimes the solution is complex, involving complex numbers (2 complex), sometimes they are real and distinct (2 real), and sometimes they are real and coincident (still two real, but they are the same).

In the case of

x^2+3x = 3, or

x² + 3x -3 = 0

we apply the quadratic formula to get

x =  (-3 +/- sqrt(3^2+4(1)(3))/2

to give the two solutions

{(sqrt(21)-3)/2, -(sqrt(21)+3)/2,}

both of which are real.

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