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Law Incorporation [45]
3 years ago
3

40 POINTS

Mathematics
2 answers:
Anna71 [15]3 years ago
8 0

we know that

The area of a rectangle is equal to

A=L*W

where

L is the length side of the rectangle

W is the width side of the rectangle

The perimeter of the rectangle is equal to

P=2L+2W

Step 1

Find the dimensions of the rectangles

we know that

the distance between two points is equal to the formula

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance A-F

A-F is the width of the rectangles ABEF and ACDF

A(0,6)\\F(5,2)

substitute the values in the formula

dAF=\sqrt{(2-6)^{2}+(5-0)^{2}}

dAF=\sqrt{(-4)^{2}+(5)^{2}}

dAF=\sqrt{41}\ units

so

W=\sqrt{41}\ units

Find the distance F-E

F-E is the length side of the rectangle ABEF

F(5,2)\\E(11,10)

substitute the values in the formula

dFE=\sqrt{(10-2)^{2}+(11-5)^{2}}

dFE=\sqrt{(8)^{2}+(6)^{2}}

dFE=10\ units

Find the distance F-D

F-D is the length side of the rectangle ACDF

F(5,2)\\D(14,14)

substitute the values in the formula

dFD=\sqrt{(14-2)^{2}+(14-5)^{2}}

dFD=\sqrt{(12)^{2}+(9)^{2}}

dFD=15\ units  

Step 2

the dimensions of the rectangle ABEF are

L=10\ units\\W=\sqrt{41}\ units

Find the area of the rectangle ABEF

A=10*\sqrt{41}=10\sqrt{41}\ units^{2}

A=64.03\ units^{2}

Step 3

the dimensions of the rectangle ACDF are

L=15\ units\\W=\sqrt{41}\ units

Find the area of the rectangle ACDF

A=15*\sqrt{41}=15\sqrt{41}\ units^{2}

A=96.05\ units^{2}

Step 4

The ratio of the area of rectangle ABEF to the area of rectangle ACDF is

(10\sqrt{41}/15\sqrt{41})=(10/15)=2/3

Step 5

Find the perimeter of the rectangle BCDE

we know that

W=\sqrt{41}\ units

the length of the rectangle BCDE is equal to the distance E-D

ED=FD-FE

ED=15-10=5\ units

L=5\ units

P=2*5+2*\sqrt{41}=22.81\ units

therefore

the answer part a) is

2:3

the answer part b) is

the area of the rectangle ABEF is 64.03\ units^{2}

the answer part c) is  

the area of the rectangle ACDF is 96.05\ units^{2}

the answer Part d) is  

the perimeter of the rectangle BCDE is 22.81\ units



Murrr4er [49]3 years ago
5 0

Answer:

1.) 2:3

2.) 64.03

3.) 96.05

4.) 22.81

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Juan is three times as old as Gabe and Gabe is six years older than Catherine. If the sum of their ages is 149, how old is each
cestrela7 [59]

Answer:

Juan = 93 years.

Gabe = 31 years.

Catherine = 25 years.

Step-by-step explanation:

Let the age of Juan = J

Let the age of Gabe = G

Let the age of Catherine = C

<em>Translating the word problem into an algebraic equation, we have;</em>

J = 3G ..........equation 1

G = C + 6 ........equation 2

J + G + C = 149 ........equation 3

<em>We would solve the linear equations by using the substitution method; </em>

<em>Substituting equation 2 into equation 1;</em>

J = 3(C + 6)

J = 3C + 18 ........equation 4

<em>Substituting equation 2 and equation 4 into equation 3;</em>

(3C + 18) + (C + 6) + C = 149

<em>Simplifying the equation, we have;</em>

5C + 24 = 149

5C = 149 - 24

5C = 125

C = \frac {125}{5}

C = 25 years.

To find G; from equation 2

G = C + 6

Substituting the value of "C" into equation 2, we have;

G = 25 + 6

G = 31 years.

To find J; from equation 1

J = 3G

Substituting the value of "G" into equation 1, we have;

J = 3 * 31

J = 93 years.

<em>Therefore, Juan is 93 years old, Gabe is 31 years old and Catherine is 25 years old. </em>

4 0
3 years ago
Given the supply and demand functions below, find the demand when p = $12.
goblinko [34]
S(p)= 5(12)= $60
D(p)= 120-4(12)= $72
3 0
3 years ago
Read 2 more answers
Examplelt: A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall
harina [27]

Answer :

  • 4167 bricks.

Explanation :

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here,

{\qquad \dashrightarrow{ \sf{Length=10 \: m=1000 \: cm}}}

\qquad \dashrightarrow{ \sf{Thickness=24 \: cm}}

\qquad \dashrightarrow{ \sf{Height=4 m=400 \: cm}}

Therefore,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: the  \: wall = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: the  \: wall = 1000 \times 24 \times 400 \:  {cm}^{3} }}}

Now, each brick is a cuboid with Length = 24 cm, Breadth = 12 cm and height = 8 cm.

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: each  \: brick = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: each  \: brick = 24 \times 12 \times 8 \:  {cm}^{3} }}}

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: bricks  \: required =  \dfrac{volume \: of \: the \: wall}{volume \: of \: each \: brick} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =  \dfrac{1000 \times 24 \times 400}{24 \times 13 \times 8} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =   \bf \: 4166.6} }}

Therefore,

  • <u>The wall requires 4167 bricks. </u>

6 0
2 years ago
Bob has $1.15 in his wallet which contains 18 coins in nickels and dimes. Find how many nickles he has.​
ratelena [41]

9514 1404 393

Answer:

  13 nickels

Step-by-step explanation:

Let n represent the number of nickels Bob has. Then 18-n is the number of dimes, and the total value (in cents) is ...

  5n +10(18-n) = 115

  -5n +180 = 115 . . . simplify

  65 = 5n . . . . . add 5n-115

  13 = n . . . . . . . divide by 5

Bob has 13 nickels.

5 0
3 years ago
Read 2 more answers
Question is download
klemol [59]

The value of the digit in the tenth place is <u>10 times</u> as much as the digit in the hundredth place.

<h3>What is a digit?</h3>

A digit is a symbol used to construct numerals from 0 to 9, used to represent numbers.

We use the numerals such as 0, 1, 2, 3,4, 5,6,7,8, and 9 to form digits that represent a combination of numbers and do arithmetic operations.

All numerals are made up of the ten digits listed above.

Thus, we can conclude that the value of the digit in the tenth place is <u>10 times</u> as much as the digit in the hundredth place.

Learn more about digits at brainly.com/question/26856218

#SPJ1

4 0
2 years ago
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