The amount of rhubarb in the original recipe is 31\2 cups. Using what you know of whole numbers and what you know of fractions, explain how you could triple that mixed number

The object will take 2.5 seconds to reach its maximum height. The maximum height attained by the object will be 42.375 meters.

To find how long it takes for the object to reach its maximum height, we can use the projectile motion formula

h(t) = -4.9t² + v0t + h0.

The maximum height is achieved when the vertical velocity component is zero. This happens midway through the flight.

The time to reach maximum height can be found using the formula

t = v0 / g,

where v0 is the initial velocity and

g is the acceleration due to gravity (9.8 m/s²).

In this case, v0 = 24.5 m/s and g = 9.8 m/s², so the time (t) to reach maximum height is

t = 24.5 m/s / 9.8 m/s² = 2.5 s.

The maximum height (H) can be calculated as the initial height plus the displacement due to the initial velocity upwards before it starts coming down.

Since we know the time to reach the maximum height, we can plug it into the original equation h(t) to find the maximum height:

H = -4.9(2.5)² + 24.5(2.5) + 11

We then find that

H = 42.375 m.

divide 630 by 18

630/18 = 35

so unit rate would be 35 meters per second

The length of the given rectangular field is 120 yd while the width is 70 yd.

A rectangle is a geometrical figure in which opposite sides are equal.

The angle between any two consecutive sides will be 90 degrees.

Area of rectangle = length × width.

Perimeter of rectangle = 2( length + width).

Let's say the length of a rectangle is L while the width is W

So,

Perimeter 2(L + W) = 380

And,

L = W + 50

By substitution,

W + 50 + W = 190

2W + 50 = 190

2W = 140

W = 70

And

L = 70 + 50 = 120

Hence "The length of the given rectangular field is 120 yd while the width is 70 yd.

For more about rectangles,

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Follow below steps:

The location of the oil tanker and the luxury liner at time t hours after 2 p.m.:

The oil tanker's location: (-14t, 0)

The luxury liner's location: (0, 16t)

The distance d between the oil tanker and the luxury liner at time t:

The distance formula = √((-14t - 0)^2 + (0 - 16t)^2) = √(196t^2 + 256t^2) = √(452t^2) = 2√(113)t

The distance d between the oil tanker and the luxury liner at that time is [tex]\( \sqrt{452} \cdot t \)[/tex] kilometers.

To solve this problem, we'll use the concepts of distance, speed, and coordinate geometry.

Let ( t ) be the number of hours after 2 p.m.

Location of the Oil Tanker:

At ( t ) hours after 2 p.m., the oil tanker has traveled ( 14t ) kilometers west from the origin.

So, its location is  (-14t, 0) .

Location of the Luxury Liner:*

At ( t ) hours after 2 p.m., the luxury liner has traveled ( 16t ) kilometers north from the origin.

So, its location is  (0, 16t).

Distance between the Oil Tanker and the Luxury Liner:

We'll use the distance formula: [tex]\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).[/tex]

[tex]\[ d = \sqrt{(-14t - 0)^2 + (0 - 16t)^2} \][/tex]

[tex]\[ d = \sqrt{(-14t)^2 + (-16t)^2} \][/tex]

[tex]\[ d = \sqrt{196t^2 + 256t^2} \][/tex]

[tex]\[ d = \sqrt{452t^2} \][/tex]

[tex]\[ d = \sqrt{452} \cdot t \][/tex]

So, the location of the oil tanker is [tex]\( (-14t, 0) \)[/tex], the location of the luxury liner is [tex]\( (0, 16t) \)[/tex], and the distance between them at ( t ) hours after 2 p.m. is [tex]\( \sqrt{452} \cdot t \)[/tex] kilometers.

cylinder B is 4 times the size of A

Answer:

375

Step-by-step explanation:

how to equate: to add 12.8 to -15 subtract 12.8 from 15

15-12.8 =2.2

 since 15 is negative and is larger than 12.8 the answer would be negative so 2.2 becomes -2.2

-15 + 12.8 = -2.2

-15 + 12.8 equates to -2.2 by subtracting the smaller absolute value from the larger one and considering the sign of the number with the larger absolute value. The solution involves basic arithmetic with negative and positive numbers.

Solving the Equation -15 + 12.8

To solve the equation -15 + 12.8, follow these steps:

So, -15 + 12.8 equates to -2.2.

Gwen will owe her parents a total of $309.

To represent the total amount Gwen owes her parents after borrowing $300 with 3% interest over a year, we can use the formula for simple interest:

[tex]\[ \text{Total amount} = \text{Principal} + \text{Interest} \][/tex]

Given:

Principal (P) = $300

Interest rate (r) = 3% or 0.03 (as a decimal)

Time (t) = 1 year

Now, plug the values into the formula:

[tex]\[ \text{Interest} = P \times r \times t \][/tex]

[tex]\[ \text{Interest} = 300 \times 0.03 \times 1 \][/tex]

[tex]\[ \text{Interest} = 9 \][/tex]

Now, add the interest to the principal to get the total amount:

[tex]\[ \text{Total amount} = 300 + 9 \][/tex]

[tex]\[ \text{Total amount} = \$309 \][/tex]

So, the equation to represent the total amount Gwen owes her parents is:

[tex]\[ \text{Total amount} = 300 + 300 \times 0.03 \times 1 \][/tex]

[tex]\[ \text{Total amount} = 300 + 9 \][/tex]

[tex]\[ \text{Total amount} = \$309 \][/tex]

Therefore, Gwen will owe her parents a total of $309.

If c – 8 is an odd integer, then c could be any even number greater than 8 such as 10, 12, 14, 16, because when you subtract 8 from these even numbers, the result is an odd integer.

The question requires us to find the value of c if c : 8 is an odd integer. So, if c : 8 is an odd integer, that means whatever number c is, when you subtract 8 from it, you should get an odd number. Odd numbers are integers that are not divisible by 2 and they usually end in 1, 3, 5, 7, or 9. Therefore, the possible values for c can be any even number greater than 8, such as 10, 12, 14, 16, and so on. That's because when you subtract 8 from these even numbers, the result will be an odd integer.

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To solve this problem, it is necessary determine the integrations limits first.

The solution is:

V = 1666,13×π vol units

In order to express the information in spherical coordinates, we have:

x = ρ×cosθ×sinφ

y = ρ×sinθ×sinφ

z = ρ×cosφ

and     dV = ρ²×sinφ×dρ×dθ×dφ

0 ≤ φ ≤ π

0 ≤ θ ≤ 2×π

x² + y² + z² = 25       and         x² + y² + z² = 1    are two concentric spheres with center at the origin and radius 5 and 1 respectevely .

According to that:

1 ≤  ρ ≤ 5

V = ∫∫∫ ( x² + y² )× dV  

V  =   ∫∫∫ ( x² + y² )×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ (  ρ²×cos²θ×sin²φ  + ρ²×sin²θ×sin²φ ) ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [  ρ²×sin²φ ( cos²θ + sin²θ )  ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [  ρ²×sin²φ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [ ρ⁴×sin³φ×dρ×dθ×dφ

V =  ∫₁⁵ρ⁴×dρ ×∫dθ ×∫sin³φ×dφ

V =  ρ⁵/5 |₁⁵×∫dθ ×∫sin³φ×dφ

ρ⁵/5 |₁⁵  = 5⁵/5 - 1⁵/5   = 3125 / 5 - 1/5  = 3124/5

V = 3124/5× θ |₀²π×∫sin³φ×dφ

θ |₀²π  = 2×π - 0

V = 3124/5× (2×π) ×∫sin³φ×dφ

∫sin³φ×dφ  [ 0  y π ] = 4/3   (from Simbolab)

V = 3124/5× (2×π) ×4/3

V = 1666,13×π   vol units

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