All categories

2 years ago

HELP IS NEEDED FAST PLEASE

Mathematics

1 answer:

vfiekz [6]2 years ago

7 0

Answer:

17.1

Step-by-step explanation:

If you have 15.50 and you divide 0.75 you get 20.6 and if you subtract 3.50 you get 17.1 which is your answer. Boom

You might be interested in

Which coordinate plane shows the graph of the function y = ?

Brilliant_brown [7]

Y=1/2 x so graph B i think.

5 0

3 years ago

Read 2 more answers

What is a 1/4 of ?/100

irakobra [83]

$\frac{1}{4}\cdot100=25$
......

8 0

3 years ago

Read 2 more answers

A study conducted to determine the relationship between the age, x, in years of a certain brand of motorcycle and its value, y,

kirza4 [7]

Based on the equation, y = -750·x + 8500, the estimated value of a

motorcycle that is 5 years old is <u>$4,750</u>.

<h3>How can the given equation help to estimate the value of the motorcycle?</h3>

The regression equation that models the data is; y = -750·x + 8500

Where;

y = The value of the brand of motorcycle in dollars

x = The age of the motorcycle

When the motorcycle is 5 years old, we have;

x = 5

y = -750 × 5 + 8500 = 4750

Based on the equation, the estimated value of a motorcycle that is 5 years old, y =<u> $4,750</u>

Learn more equations here:

brainly.com/question/20475360

4 0

2 years ago

brian makes an investment in his company’s stock . what does the stock chart say about the stock price today? The down arrow ind

Ivan

Answer:50$ and 2

Step-by-step explanation:

3 0

2 years ago

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

Now, y = vx ⇒ v = y/x

$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

5 0

3 years ago