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Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

5 0

Answer:

literally at this point i think its (D)

Step-by-step explanation:

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I need help now it's an emergency please help me with this

marshall27 [118]

Answer:

Population of the 48th generation will be 4469.

Step-by-step explanation:

Recursive formula by which the population is increasing,

$L_{n+1}=L_n+95$

L₀ = 4

Common difference 'd' = 95

Recursive formula represents a linear growth in the population.

Therefore, explicit formula for the given sequence will be,

$L_n$ = L₀ + (n - 1)d [Explicit formula of an Arithmetic sequence]

Here n = Number of terms

L₄₈ = L₀ + (48 - 1)(95)

= 4 + 4465

= 4469

Therefore, population of the 48th generation will be 4469.

4 0

3 years ago

The quotient of 3,200 and 4 will have -blank- zeros.

Nostrana [21]

B. It will have two zeros. Hope this helps!

3 0

3 years ago

Read 2 more answers

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$