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3 years ago

A park has an area of 12.5 sq miles and a width of 5 miles. What is the perimeter of the park?

Mathematics

2 answers:

Andrews [41]3 years ago

5 0

Answer:

Step-by-step explanation:

12.5/5=2.5

2(2.5)+2*5=

5+10=15

djverab [1.8K]3 years ago

3 0

Answer:

actually not sure if it's correct.

2.5 sq miles??

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(45 1/5)- (-16 9/11)

Viktor [21]

On the TI-34 multiview the answer was
62 1/55

7 0

4 years ago

Find the surface area of the prism.<br> 10 ft length<br> 1 ft height<br> 5 ft width

Elenna [48]

Answer:

Surface Area = 130 sq ft

Step-by-step explanation:

If you don't have a picture, draw a picture so you can see what's up. See image. Surface area is the area of all the faces of a 3d shape (a solid) all added up. This solid is a box. It has six faces. Each face is a rectangle. The top and bottom are the same. The front and back are the same. The left and right sides are the same. See image.

Area of a rectangle is:

A = l × w

Area of the top:

5 × 10 = 50

Area of the bottom also 50.

Area of the left side:

1 × 5 = 5

Area of the right side is also 5.

Area of the front:

1 × 10 =10

Area of the back is also 10.

ADD up all 6 areas:

50+50+5+5+10+10

Surface area = 130

Area has square units.

SA = 130 square ft

3 0

2 years ago

F(x)=4x^2-17x+3 how many distinct real number zeros does f have?

Harrizon [31]

Answer:

Step-by-step explanation:

2 distinct real number zeros

Step-by-step explanation:

The discriminant b^2 - 4ac

= (-17)^2 - 4*4*3

= 241

So it has 2 distinct roots.

4 0

3 years ago

merlin worked these hours during the week: 4.5, 8.75, 9.5, 9, and 4.25 hours. what was his gross pay for the week, if he makes $

Usimov [2.4K]

Hello,

as 4.5+8.75+9.5+9+4.25=36 and 36*9.78=352.08

King'Arthur must pay him $352.08

5 0

3 years ago

Read 2 more answers

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

3 years ago