The price of milk increased from $2 to $2.50. What is the percent of change for the price of milk?

Please Help!

Tamiku [17]

Answer:

n

Step-by-step explanation:

the answer is -60, which is not an option

4 0

2 years ago

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a fun

MrMuchimi

Answer:

$y{x} = \sqrt{7+2Inx}$

Step-by-step explanation:

$y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}$

Let say; By y(x)= y(e)

we have;

$y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0$

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

$y^{1} (x) = 0 + 1/ xy$

$y^{1} = 1/xy$

dy/dx = 1/xy

$\int\limits {y} \, dxy = \int\limits \, dx/x$

$y^{2}/2 Inx + C$

RECALL: y(e) = 3

$(3)^{2} / 2 = In (e) + C$

$\frac{9}{2} =In(e)+C$

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

$y^{2} = {7+2Inx}$

$y{x} = \sqrt{7+2Inx}$

8 0

3 years ago

Bashir works in a weather station and has launched a weather ballon. The height of the baloon is 4 kilometers and it is 1 kilome

andrew-mc [135]

Answer:

(c) 4.1 km

Explanation:

We are given that the height of the balloon is 4 km and that it is 1 km away from the station.

Setting up the diagram that describes the given scenario, we would end up with a right-angled triangle as the one shown in the attached image

where:

height of the balloon = 4 km represents one side of the triangle

distance between the base of the balloon and the station = 1 km represents the other side

distance between the balloon and the station is the length we want to find and it represents the hypotenuse of the triangle.

To get the length of the hypotenuse, we will use the Pythagorean theorem

$hypotenuse = +\sqrt{(side1)^2 + (side2)^2}$

Substitute with the givens:

$hypotenuse = +\sqrt{(4)^2 + (1)^2}$ = √17 ≈ 4.1 km

Hope this helps :)

3 0

2 years ago

2.

abruzzese [7]

Answer:

a) 30 kangaroos in 2030

b) decreasing 8% per year

c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo

Step-by-step explanation:

We assume your equation is supposed to be ...

P(t) = 76(0.92^t)

__

a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30

In the year 2030, the population of kangaroos in the province is modeled to be 30.

__

b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.

The population is decreasing by 8% each year.

__

c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.

P(100) = 75(0.92^100) = 76(0.0002392)

P(100) ≈ 0.0182, about 1/55th of a kangaroo

5 0

3 years ago

Me dicen la sifra 5p-14=8p+19

Alex

5p-14=8p+19
8p-5p = -3p
-3p-14=18
14+18 = 33
33/-3=-11

7 0

2 years ago