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2 years ago

3 ½ + 1 1/3 help me out I’ll give you v bucks

Mathematics

2 answers:

AnnyKZ [126]2 years ago

6 0

7 1/6 is the answer
Hope this help!
Fortnight is garbage, no one wants your v bucks

ratelena [41]2 years ago

5 0

Answer:

(3 * (1 / 2)) + 1 1/3 = 2.83333333

Step-by-step explanation:

Hope this answer helps you :)

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Some please help!!!...................

sukhopar [10]

Answer:

should be 5, but seems too easy tbh

5 0

3 years ago

Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925

Svetach [21]

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h$

Carry out the binomial expansion in the numerator:

$(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4$

Then the 9⁴ terms cancel each other, so in the limit we have

$\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h$

Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:

$\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)$

Then when h converges to 0, each remaining term containing h goes to 0, leaving you with

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}$

or choice C.

Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:

$f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h$

We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.

8 0

3 years ago

A line of best fit predicts that when x equals 35, y will equal 34.785, but y

Dafna1 [17]

Answer:

Step-by-step explanation:

4 0

3 years ago

What will be the amount in an account with initial principal $9000 il interest is compounded continuously at an annual rate of 3

Sindrei [870]

Answer:

$11,258.30

Step-by-step explanation:

6 0

3 years ago

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