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3 years ago

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Jim wants to build a toy house. He plans for the width to be 2 feet and the length to be 3 feet. Then he decides to add to the l

ength. The amount he wants to add to the length is shown by the variable y .

1 answer:

Cloud [144]3 years ago

4 0

Lets w width, l length, y amount add, A area, P perimeter

A=lw=(3+y)(2)=6+2y ft²

p= 2l+2w= 2(3+y)+2(2)= 6+2y+4 =10+2y

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Please answer correctly !!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!

posledela

Answer: 13x-1

Step-by-step explanation:

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3 years ago

Explain how to determine the number of square feet required to cover your rectangular garden that has a length of 10 feet and a

abruzzese [7]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Denise and Jacqueline are training for a swimming and running race.

bezimeni [28]

Answer:

Jacqueline runs at 6.25 minutes per mile

Step-by-step explanation:

Here we have that the total training time for Jacqueline = 54 minutes

The time for which Jacqueline swims = 30 minutes

Therefore, the time in which Jacqueline runs = 54 min. - 30 min. = 20 minutes

Therefore the equation for Jacqueline's speed = Distance/Time we have

Jacqueline's pace = 3.2 miles/20 minutes = 0.16 miles/minute

Therefore Jacqueline's pace = 0.16 miles/minute

The equation for Jacqueline's pace in miles/minute is presented as follows

Jacqueline's pace in miles/minute = Time of running/Distance

Jacqueline's pace in miles/minute = 1/0.16 miles/minute = 6.25 minutes/mile.

5 0

3 years ago

Josh is building a sandbox for his little sister. He uses these plans, which have a scale of 2 in. : 3 ft.

skelet666 [1.2K]

Answer:

Width of 6ft and length of 9ft

Step-by-step explanation:

The sandbox has a ratio of 2:3. If you plug the numbers in with the ratio, you can find the actual dimensions.

Let x be the length

Let y be the width

(Length)

2in / 3ft = 6in / (x)ft

2in / 3ft = 6in / <u><em>9ft </em></u>

(Width)

2in / 3ft = 4in / (y)ft

2in / 3ft = 4in / <em><u>6ft</u></em>

6 0

3 years ago

sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = \left\{ \begin{array}{ll} 2\cos(\pi x) \text{ for } x \leq -1 \\ \\ \displaystyle \frac{2}{\cos(\pi x)}\text{ for } x > -1 \end{array} \right.$

And we want to find:

$\displaystyle \lim_{x\to -1}f(x)$

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

$\displaystyle \lim_{x\to-1^-}f(x)$

Since we are approaching from the left, we will use the first equation:

$\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)$

By direct substitution:

$=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2$

Right-Hand Limit:

$\displaystyle \lim_{x\to -1^+}f(x)$

Since we are approaching from the right, we will use the second equation:

$=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}$

Direct substitution:

$\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2$

So, we can see that:

$\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2$

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

$\displaystyle \lim_{x \to -1}f(x)=-2$

Our answer is A.

8 0

3 years ago