Use the figure to write and solve an equation for x

A used-droid dealership buys a droid for $2900 and then sells it for $4300. What is the percent increase?

Anna71 [15]

The percent increase is 48.28

Step-by-step explanation:

Given,

Purchase price of used-droid = $2900

Selling price of used=droid = $4300

Profit = Selling price - purchase price

Profit = 4300-2900 = $1400

Percent increase = $\frac{Profit}{Purchase\ price}*100$

$Percent\ increase=\frac{1400}{2900}*100\\\\Percent\ increase=\frac{140000}{2900}\\\\Percent\ increase=48.275\%$

Rounding off to nearest hundredth;

Percent increase = 48.28%

The percent increase is 48.28

Keywords: percentage, subtraction

Learn more about percentages at:

#LearnwithBrainly

8 0

3 years ago

Jessica invests $250 each quarter in an ordinary annuity with a 5% interest rate compounded quarterly. Determine the accumulat

madam [21]

Answer:

Step-by-step explanation:

Usimg formula for calculating compound interest.

A = P(1+r/n)^nt

P is the principal =$250

r is the rate = 5%

t is the time = 25years

n = 1/4(compounded quarterly)

Substituting to get the amount A.

A = 250(1+5/25)^25/4

A = 250(1+0.2)^6.25

A = 250(1.2)^6.25

A = 250(3.125)

A = $781.31

Hence the accumulated amount in Jessica's annuity after 25 years is $781.31

4 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago

Please solve these for 20 points : solving formulas for a variable

Digiron [165]

6. (A/pi = r^2)
7. [(P - 2l)/2 = w]
8. [C/(2pi)= r]
9. (2A/h = b)
10. (E/c^2 = m)

6 0

3 years ago

Can somebody help me please

Alexus [3.1K]

Answer:

Step-by-step explanation:

Y=7 enjoy

4 0

3 years ago

Read 2 more answers