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solmaris [256]
2 years ago
6

Dario increased his savings from $350 to $413. While Monica increased her savings from $225 to $270. Who had a higher percent of

increased in savings. The difference in the percents of increase is
Mathematics
1 answer:
Anettt [7]2 years ago
4 0
The percentage in increase can be calculated as follows:
% of increase = (difference between new and old amounts / old value) *100

For Dario:
difference in values = 413 - 350 = 63
% of increase = (63/350)*100 = 18%

For Monica:
difference in values = 270 - 225 = 45
% of increase = 20%

Therefore,
Monica has higher percent of increase.
The difference in percents = 20% - 18% = 2%
You might be interested in
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
2 years ago
You sell sporting goods. Your wages depend on the value of your sales. One week you sold $3,500 in sporting goods, earning $950.
Semmy [17]

Answer:

y = 0.2x + 250

Step-by-step explanation:

let the sales be x and y be earnings

thus,

given

x₁ = $3,500 ; y₁ = $950

and,

x₂ = $2,800 ; y₂ = $810

Now,

the standard line equation is given as:

y = mx + c

here,

m is the slope

c is the constant

also,

m = \frac{y_2-y_1}{x_2-x_1}

or

m = \frac{810-950}{\textup{2,800-3,500}}

or

m = 0.2

substituting the value of 'm' in the equation, we get

y = 0.2x + c

now,

substituting the x₁ = $3,500 and y₁ = $950 in the above equation, we get

$950 = 0.2 × $3,500 + c

or

$950 = $700 + c

or

c = $250

hence,

The equation comes out as:

y = 0.2x + 250

7 0
1 year ago
A car traveled at a constant speed. The graph shows how far the car traveled, in miles, during a given amount of time, in hours.
snow_tiger [21]

Answer:

Step-by-step explanation:

A car traveled at a constant speed as shown in the graph.

Distance traveled is on the y-axis and duration of travel on the x-axis.

Point A(3.5, 210) shows,

Distance traveled = 210 miles

Time to travel = 3.5 hours

So the point (3.5, 210) shows the distance traveled by the car in 3.5 hours is 210 miles.

Slope of the line = speed of the car = \frac{\text{Distance traveled}}{\text{Time}}

                           = \frac{210}{3.5}

                           = 60 mph

Now we will find the speed of the car at another point B(1, 60).

If the speed of car is same as the point B as of point A, point B will lie on the graph.

Speed of the car at B(1, 60) = \frac{\text{Distance traveled}}{\text{Time}}

                                             = \frac{60}{1}

                                             = 60 mph

Hence, we can say that point B(1, 60) lies on the graph.

8 0
1 year ago
What is the difference between 76.82 and 2.761
Sophie [7]

74.059 is the answer.




Hope this helps...


3 0
1 year ago
Read 2 more answers
I WILL AWARD BRAINLIEST!! PLEASE HELP!!! The figure below shows the movement of a pedestrian from point B to point E. Using the
MissTica

A) The speed of the pedestrian BC is 5 km/h

    The speed of the pedestrian CD is 0 km/h

    The speed of the pedestrian DE is 5 km/h

B) He arrived E since the stop after 6 hours

C) The formula for section BC is d(t) = 40 - 5t

    The formula for section CD is d(t) = 20

    The formula for section DE is d(t) = 50 - 5t

Step-by-step explanation:

A)

In the time-distance graph the speed is the rate of change of distance

and that mean speed = Δd/Δt ⇒ (slope of the line)

In line BC:

1. Δd = 40 - 20 = 20 km

2. Δt = 4 - 0 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of the pedestrian BC is 5 km/h

In line CD:

1. Δd = 20 - 20 = 0 km

2. Δt = 6 - 4 = 2 hours

3. The speed = 0 ÷ 2 = 0 km/h

The speed of the pedestrian CD is 0 km/h

In line DE:

1. Δd = 20 - 0 = 20 km

2. Δt = 10 - 6 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of the pedestrian DE is 5 km/h

B)

∵ He stop at t = 4 hours

∵ He arrived at point E at t = 10 hours

∵ 10 - 4 = 6 hours

He arrived E since the stop after 6 hours

C)

The speeds are represented by lines

The form of the equation of a line is f(x) = mx + c, where m represents

the slope of the line and c is the y-intercept (y when x = 0)

1. f(x) is d(t)

2. m is the speed

3. x is t

4. You can find c by substitute d and t by any point on the line

Line BC

Line BC has negative slope because d decreases when t increases

∵ m = -5 and c = 40

∴ d(t) = 40 - 5t

The formula for section BC is d(t) = 40 - 5t

Line CD

Line CD is a horizontal line (equation any horizontal line is y = c)

∴ m = 0 and c = 20

∴ d(t) = 20

The formula for section CD is d(t) = 20

Line DE

Line DE has negative slope because d decreases when t increases

∵ m = -5

∴ d(t) = -5t + c

To find c substitute the coordinates of point D in the equation

∵ The coordinates of point D are (6 , 20)

∴ 20 = -5(6) + c

∴ 20 = -30 + c

Add 30 to both sides

∴ c = 50

∴ d(t) = 50 - 5t

The formula for section DE is d(t) = 50 - 5t

Learn more:

You can learn more about the distance, speed, and time in

brainly.com/question/5102020

#LearnwithBrainly

3 0
1 year ago
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