A campground has cabins that can each hold 28 campers. There are 148 campers that are visiting the campground. How many cabins are full if there are 28 campers in each cabin? Please help i'm desperate and i'm marking most brainliest.

Answer: [tex]30\ ft^2[/tex]

Step-by-step explanation:

We are given that

The base of the rectangle = 2 yards

The height of the rectangle = 5 feet

We know that

1 yard = 3 feet

Then , the base of the rectangle = [tex]2 \times3=6\text{ feet}[/tex]

Now, the are of the rectangle will be :-

[tex]\text{Area}=\text{Height x Base}\\\\\Rightarrow\ \text{Area}=5\times6\text{ square feet}\\\\\text{Area}= 30\text{ square feet}[/tex]

Hence, the area of the rectangle = [tex]30\ ft^2[/tex]

Answer:

i think is 30ft2

Step-by-step explanation:

Answer: 0.12

Step-by-step explanation:

Answer:

3/4.

Step-by-step explanation:

4/8 simplified=1/2 6/8 simplified=3/4.

The sentence 'The difference of Rhonda's height and 16 is 52' translates into the equation 'h - 16 = 52'. Here, 'h' represents Rhonda's height.

The subject matter of this question falls under Mathematics. Specifically, it deals with basic algebra where you have to translate word problems to an algebraic equation. Remember that the phrase 'the difference of' in math refers to the operation of subtraction. So, when you read 'the difference of Rhonda's height and 16 is 52,' that means Rhonda's height (which we’ll denote as 'h') minus 16 equals 52 (52 comes from the phrase 'is 52' because 'is' usually represents equal '='). Therefore, when this sentence is written in the form of an equation, it will look like: h - 16 = 52.

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The donut and bagel timers at Carlie's bakery will buzz together every 2 hours, so they will buzz at the same time for the first time at 10:00 a.m.

This question is about finding the Least Common Multiple (LCM) of the two numbers - 30 minutes (the time the donut timer buzzes) and 40 minutes (the time the bagel timer buzzes). The LCM of 30 and 40 is 120 minutes. This means that the timers for the bagels and the donuts will buzz at the same time every 120 minutes, or 2 hours. Since Carlie starts both timers at 8:00 a.m., the timers will buzz together for the first time at 10:00 a.m.

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Answer:

1

Step-by-step explanation:

GCF(18, a, 20ab) = 1

18 and a do not necessarily have any factor in common, so their GCF is 1. Adding more terms to the list will not increase the GCF beyond that value.

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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The equation for the given situation is y=48+18x.

Given that, a ski club charges a $48 membership fee, plus $18 to rent ski equipment per day.

In mathematics, an equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.

Given that, equipment is renting for x days.

Let the total cost of membership at the club be y.

Now, y=48+18x

Therefore, the equation for the given situation is y=48+18x.

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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.