All categories

2 years ago

Lauren bought 4 bags of popcorn for $3.00. What is the unit rate per bag of popcorn."?

Mathematics

1 answer:

S_A_V [24]2 years ago

7 0

Answer:

$0.75

Step-by-step explanation:

Given that

Number of bags of popcorn bought = 4

Total money spent = $3.00

To find:

Unit rate per bag of popcorn = ?

i.e. price of one bag of popcorn is to be find out.

Solution:

We can use ratio here to find the rate of one bag of popcorn.

4 bags bought at $3

4 bags : $3

Let us divide both the sides with 4.

$\frac{4}4$ bags : $ $\frac{3}4$

1 bag bought at $ 0.75

We can alternatively use unitary method.

4 bags are bought at $ 3

1 bag is bought at $ $\frac{3}{4}$

1 bag is bought at $0.75.

So, unit rate per bag of popcorn is $0.75.

You might be interested in

Which graph matches the equation y+3=2(x+3)?

Vlad1618 [11]

Answer:

The graph that includes points (-3,-3) and (0,3)

Step-by-step explanation:

In the pictures attached, the options are shown.

The equation:

y+3=2(x+3)

has the point-slope form, which is:

y-y₁=m(x-x₁)

where (x₁, y₁) is a point on the line and m is its slope. This means that (-3,-3) is on the line. To know the y-intercept of the line, we have to replace x = 0 into the equation, as follows:

y+3=2(0+3)

y+3 = 6

y = 6 - 3

y = 3

Then, point (0, 3) is on the line.

8 0

2 years ago

Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist

GaryK [48]

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

$l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}$

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

$l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}$

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile = 1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

$l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}$

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

5 0

2 years ago

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:

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

Isoke is solving the quadratic equation by completing the square.10x2 + 40x – 13 = 0 10x2 + 40x = 13 A(x2 + 4x) = 13What is the

Elina [12.6K]

we have

$10x^{2} + 40x - 13 = 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$10x^{2} + 40x =13$

Factor the leading coefficient

$10(x^{2} + 4x) =13$ ----------> the value of A is $10$

Complete the square. Remember to balance the equation by adding the same constants to each side

$10(x^{2} + 4x+4) =13+40$

$10(x^{2} + 4x+4) =53$

Rewrite as perfect squares

$10(x+2)^{2} =53$

$10(x+2)^{2} =53 \\ (x+2)^{2} =\frac{53}{10} \\ \\ x+2=(+/-)\sqrt{\frac{53}{10}} \\ \\ x1=-2+\sqrt{\frac{53}{10}}\\ \\ x2=-2-\sqrt{\frac{53}{10}}$

therefore

the answer is

the value of A is $10$

3 0

2 years ago

Read 2 more answers

Given: Quadrilateral ABCD is a kite.

Leni [432]

Answer:

Step-by-step explanation:

One

The definition of a kite.

Two

Perpendicular. The diagonals meet at right angles

Three

I don't know if the term is still used, but when I first learned this material (many years ago) we called it the reflexive property.

Four

You found out that <AED = <CED which established that both are right angles. The problem is, you do not have enough information to prove SAS.

AAS is never a dependable reason for proving anything conguent.

HL is really your only choice.

8 0

2 years ago

Read 2 more answers