How do you solve 5 3/4 divided by 4 and show step by step

1 answer:

5 0

There are various ways in which to do this problem. I'd suggest
converting 5 3/4 into an improper fraction and then dividing that improper fraction by 4:

20+3 23 1 23
------- divided by 4 is --------- * ----- = ------ (answer)
4 4 4 16

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Integration of ∫(cos3x+3sinx)dx

Murljashka [212]

Answer:

$\boxed{\pink{\tt I = \dfrac{1}{3}sin(3x) - 3cos(x) + C}}$

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

$\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\ dx$

Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
Now , Rewrite using du and u .

$\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}}}}$