All categories

3 years ago

Someone help plz make sure it’s right and please no websites plz

Mathematics

2 answers:

MAVERICK [17]3 years ago

5 0

Hoped this helped. I learned this as well. It’s really easy. You’ll get it.

ruslelena [56]3 years ago

4 0

Answer:

Cos A = 3/5

Hope it helps you!

You might be interested in

James had 568 books to put into 9 boxes. How many boxes were in each box ?Where there any book left that needs another box?

vaieri [72.5K]

Answer:

63 books in each boxes

1 book left which need a box

Step-by-step explanation:

James has 568 books

He plans to put them into 9 boxes

Therefore the number of books n each boxes can be calculated as follows

= 568/9

63 remainder 1

Hence there would be 63 books in each boxes and 1 book would be left which need another box

6 0

3 years ago

Please answer illgive BRAINLIEST

Aleks04 [339]

Answer:

It's B. 31.79in²

8 0

2 years ago

12. Write the first 5 terms of each: a)

Free_Kalibri [48]

There’s no terms :( sorry i wanted to help

4 0

3 years ago

When invested at an annual interest rate of 4.7%, an account earned $1,290.33 of simple

lawyer [7]

Answer:

The answer is 82,361.49

Step-by-step explanation:

after simplifying the equation, the problem is 1290.33 divided by .27 times 3 - the quotient is 82,362.49 (in simplified form)

6 0

3 years ago

Read 2 more answers

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