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2 years ago

1. A small garden has a perimeter of 24 1/4 ft. If the lengths is 8 ft, what is the area of the garden.

Mathematics

1 answer:

Pavlova-9 [17]2 years ago

5 0

The area of the garden is 33 ft².

The number of tiles required to cover the floor is 17 tiles.

The area of a dog pen is 170/9 square yards.

We need to find the area and perimeter for the given questions.

<h3>What are the formulas to find the perimeter and the area of a rectangle?</h3>

The formula to find the perimeter of a rectangle=2(Length+Breadth)

The formula to find the area of a rectangle=Length×Breadth.

Given that, a small garden has a perimeter of 24 1/4 ft and the length is 8 ft.

Now, perimeter = 2(8+Breadth)

⇒24 1/4=16+2×Breadth

⇒97/4-16=2×Breadth

⇒Breadth=97-64/4×2=33/8 ft

Now, Area=8×33/8=33 ft²

Therefore, the area of the garden is 33 ft².

Given that, the floor measures 4 1/2 ft by 3 3/4 ft.

The area of a floor=Length×Breadth=9/2×15/4=135/8 ft²

Number of tiles to cover the floor=135/8=16.875

Therefore, the number of tiles required to cover the floor is 17 tiles.

Given that, a dog pen is 3 1/3 yards wide by 5 2/3 yards long.

Now, the area of a dog pen=10/3×17/3=170/9 square yards.

Therefore, the area of a dog pen is 170/9 square yards.

To learn more about the area and the perimeter visit:

brainly.com/question/11957651.

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Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.