Please help!! i don’t know how to do this please help asap.

Multiply (3x – 5y)(2x + 3y)

Ghella [55]

Answer:

6x² - xy - 15y²

Step-by-step explanation:

To multiply these two binomials together, we can use FOIL. FOIL stands for:

First: Multiply the first terms of the binomials together.

Outer: Multiply the first term of the first binomial by the second term of the second binomial.

Inner: Multiply the second term of the first binomial by the first term of the second binomial.

Last: Multiply the second (last) terms of the two binomials together.

(3x - 5y)(2x + 3y) = (3x)(2x) + (3x)(3y) + (-5y)(2x) + (-5y)(3y)

= 6x² + 9xy - 10xy - 15y²

= 6x² - xy - 15y²

I hope you find my answer helpful.

8 0

3 years ago

First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

8 0

3 years ago

Instructions:Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

lys-0071 [83]

(2x-1)(7xy+2) there you go hope this helps

6 0

2 years ago

Read 2 more answers

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, i

Doss [256]

C. Hope this helps :)

5 0

2 years ago

Math workshops and final exams: The college tutoring center staff are considering whether the center should increase the number

oksian1 [2.3K]

Answer:

Option D

Step-by-step explanation:

The response variable is also known as the dependent variable. It depends on another factor which is the independent variable to cause change/ response to it.

In this study, they want to test if an increase in the number of Math workshop which is the independent variable will cause a change in the final exam scores which is the dependent variable. The response to this inquiry is thus the final exam score which is also known as the response variable.

7 0

3 years ago