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3 years ago

Why is doing long division so complicated?

2 answers:

7 0

I think it is so complicated because you dont understand it to me i find it hard but once i get a few right im like oh this easy, you just need to practice because practice makes perfect. Hope this helped :)

tankabanditka [31]3 years ago

6 0

Simply put, its a LONGER way to do division. You take a longer time on showing the work.

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11111111111111111111111111111111111111111111111111111111111111111111111111111111111

kramer

Answer:

what do we have to do

Step-by-step explanation:

8 0

3 years ago

What number should be placed in the box to help complete the division calculation? Long division setup showing an incomplete cal

lianna [129]

Answer:

The unknown value is being subtracted from 226 is 160

Step-by-step explanation:

Long division setup showing an incomplete calculation

16 is in the divisor

3426 is in the dividend

2 hundreds and 1 tens is written in the quotient

3200 is subtracted from 3426 to give 226

An unknown value represented by a box is being subtracted from 226

so,

The dividend = 3426

The divisor = 16

2 hundreds means 200 and 1 tens means 10

∵ The quotient = 200 + 10 + x

∵ Dividend = divisor × quotient

∴ (16 × 200) + (16 × 10) + (16 × x) +remainder = 3426

∵ 16 × 200 = 3200

Subtract 3200 from the dividend

∴ 3426 - 3200 = 226

∵ 16 × 10 = 160

∴ 226- 160 = 66

⇒160is the unknown value

∵ 16 × x = 16x

∵ 66 - 16x = 0

∴ 66 = 16x

- Divide both sides by 16

∴ x = 4 and remainder = 2

∴ 3426 ÷ 16 = 200 + 10 + 4

∴ 3426 ÷ 16 = 214

∴ From the steps above the missing number subtracted from

226 is 160

The unknown value is being subtracted from 226 is 160

3 0

2 years ago

Let the region R be the area enclosed by the function f(x)=ln(x) and g(x)= 1/2x-2. Find the volume of the solid generated when t

Neporo4naja [7]

Answer:

$V=61.66$

Step-by-step explanation:

This problem can be solved by using the expression for the Volume of a solid with the washer method

$V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx$

where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).

Before we have to compute the limits of the integral. We can do that by taking f=g, that is

$f(x)=g(x)\\ln(x)=\frac{1}{2}x-2$

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)

x1=0.14

x2=8.21

and because the revolution is around y=-5 we have

$R=ln(x)-(-5)\\r=\frac{1}{2}x-2-(-5)\\$