A is the correct answer.......

Here is the work shown:

x^2-x-20

A is the correct answer because when using the distributive property it is the only example that results in x^2-x-20.

(x+4)(x-5)

x*x=x^2, x*-5=-5x, 4*x=4x, 4*-5=-20. Now combine all alike terms for your final answer. x^2-x-20..

please vote my answer brainliest. thanks!

5 0

3 years ago

-3 4/7 divide -1 1/14

Sedaia [141]

Answer:

hgktjeghuegkekhghkegkhetgkhekhjhkjeg

Step-by-step explanation:

8 0

3 years ago

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 .

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

6 0

3 years ago

Please please please help me!

mafiozo [28]

T(x, y) = (x - 5, y - 6) since you are trying to translate back to the original point and therefor should do the reverse set of translations

5 0

3 years ago