All categories

3 years ago

91403 divided by 21. Show work

Mathematics

2 answers:

My name is Ann [436]3 years ago

8 0

Answer:

The given Divisor = 21 and Dividend = 91403

43522191403847463110105534211

The Quotient is 4352 and the Remainder is 11

goldenfox [79]3 years ago

7 0

Answer:

The Quotient is 4352 and the Remainder is 11

Step-by-step explanation:

It's simple math

You might be interested in

what is the product of 6x – y and 2x – y 2? 8x2 – 4xy 12x y2 – 2y 12x2 – 8xy 12x y2 – 2y 8x2 4xy 4x y2 – 2y 12x2 8xy 4x y2 2y

tekilochka [14]

$(6x-y)(2x-y)=12x^{2}-8xy+y^{2}$

3 0

3 years ago

Read 2 more answers

What methods can be used to solve quadratic equations?

coldgirl [10]

The Quadratic formula

8 0

3 years ago

Read 2 more answers

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

3 0

1 year ago

The amount of methane emissions, in millions of metric

Shtirlitz [24]

Answer:

Quadratic, 711, increase

Step-by-step explanation:

I just did this problem Trust me.

6 0

4 years ago

Read 2 more answers

Help me<br> This is Geometry

sdas [7]

Step-by-step explanation:

X+3 =2X +1

X= 2

sides,,5,5 ,6 ft

perimeter = 5+5+6 = 16

4 0

3 years ago