How to write 0.624 in expanded form

Stan's lawn mower had ⅛ of a gallon of gasoline in the tank. Stand started mowing and used all of the gasoline. He put 6/10 of a

KATRIN_1 [288]

Answer:Stan used 19/40 of a gallon of gasoline.

Step-by-step explanation:

At first, Stan's lawn mower had 1/8 of a gallon of gasoline in the tank. When it got finished, he put 6/10 of a gallon of gasoline in the tank. It means that the number of gallons of gasoline that he has already put in the tank is

1/8 + 6/10 = 29/40 of a gallon.

After he mowed, 1/4 of a gallon was left in the tank. Therefore, the total amount of gasoline Stan used is

29/40 - 1/4 = 19/40 of a gallon

8 0

2 years ago

PLEASE HELP ASAP!! THIS IS LATE! :(

Svetach [21]

Answer:

States sent representatives to Congress and provided soldiers and some officers to protect the country.

4 0

2 years ago

Determine the missing exponent 2^5=1/4

Alinara [238K]

Answer:

false

Step-by-step explanation:

3 0

2 years ago

Help if x ≠ 3, how would I write x^2 - 9/ x^2 - 3x?

maria [59]

Do you want me to write it for you in standard form?

3 0

3 years ago

Separable differential equation y’ln^2y+ysqrtx=0 y(0)=e

Maksim231197 [3]

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$ .

<h3>How to solve separable differential equation</h3>

In this question we must separate each variable on each side of the equivalence, integrate each side of the expression and find an explicit expression (y = f(x)) if possible.

$\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$

$(\ln y)^{2}\,dy = -y \cdot \sqrt{x}\, dx$

$-\frac{(\ln y)^{2}}{y}\, dy = \sqrt{x} \,dx$

$-\int {\frac{(\ln y)^{2}}{y} } \, dy = \int {\sqrt{x}} \, dx$

If u = ㏑ y and du = dy/y, then:

$-\int {u^{2}\,du } = \int {x^{\frac{1}{2} }} \, dx$

$-\frac{1}{3}\cdot u^{3} = \frac{2\cdot x^{\frac{3}{2} }}{3} + C$

$u^{3} = -2\cdot x^{\frac{3}{2} } + C$

$(\ln y)^{3} = - 2\cdot x^{\frac{3}{2} } + C$

$C = (\ln e)^{3}$

$C = 1$

And finally we get the explicit expression:

$\ln y = \sqrt [3]{-2\cdot x^{\frac{3}{2} }+ 1}$

$y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$ .

To learn more on ordinary differential equations: brainly.com/question/14620493

#SPJ1

6 0

1 year ago