All categories

3 years ago

Y + 0.4= 2 Help me please PLEASE

Mathematics

1 answer:

Dmitry [639]3 years ago

3 0

Answer:

y= 1.6

You make 2 a decimal by adding .0 at the end (2.0). Then you subtract 0.4 by 2.0 which equals 1.6.

You might be interested in

Prime factors of 20 i dont know this question

dem82 [27]

Answer:

2 x 2 x 5,

Step-by-step explanation:

20 = 1 x 20

2 x 10

4 x 5

Factors of 20= 1, 2, 4, 5, 10, 20.

Prime factors are = 2 x 2 x 5,

6 0

4 years ago

Read 2 more answers

Total surface area of a solid hemisphere is a.) 3 Ir² b.) 2 Ir2 c.)4Ir2 d.)2/3 Ir3

Delicious77 [7]

Answer:

Step-by-step explanation:

8 0

4 years ago

A is what percent of b a= 2 ft b= 1 yd

dsp73

Answer:

$66\frac{2}{3}\%$

Step-by-step explanation:

Remember that

$1\ yd=3\ ft$

we have

$a=2\ ft\\b=1\ yd=1(3)=3\ ft$

In this problem

b represent the 100%

using proportion

Find out what percent of b represent a

Let

x -----> the percentage of b

$\frac{3}{100}=\frac{2}{x}\\\\x=100(2)/3\\\\x=\frac{200}{3}\%$

Convert to mixed number

$\frac{200}{3}\%=\frac{198}{3}+\frac{2}{3}=66\frac{2}{3}\%$

3 0

3 years ago

PLEASE SHOW FULL SOLUTIONS !!!!! WILL MARK BRAINLIEST FOR THE BEST ANSWER. THANK YOU AND GOD BLESS

yanalaym [24]

Answer:

System is given below

Step-by-step explanation:

Let (applicable to all three lines below)
Hard candy = x kg with price $1.60/kg
Gummy worms = y kg with price $2.20/kg
Total weight = 50 kg with mixed price $1.75/kg

Required equations:

x + y = 50 total weight
1.60x + 2.20y = 50*1.75 total price

=========================================

Note. It says don't solve but the solution below for those who is interested to know the answer.

Simplify the second equation and solve by substitution x = 50 - y:

1.6(50 - y) + 2.2y = 87.5
80 - 1.6y +2.2y = 87.5
0.6y = 7.5
y = 7.5/0.6
y = 12.5

Find the value of x:

x = 50 - 12.5 = 37.5

Hard candy = 37.5 kg and gummy worms = 12.5 kg

7 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

4 years ago