Is math a good subject?

Please help me find the area of shaded region and step by step

Bumek [7]

Answer:

Part 1) $A=60\ ft^2$

Part 2) $A=80\ cm^2$

Part 3) $A=96\ m^2$

Part 4) $A=144\ cm^2$

Part 5) $A=9\ m^2$

Part 6) $A=(49\pi -33)\ in^2$

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

$A=bh$

where

b is the base of rectangle

h is the height of rectangle

so

$A=(12)(7)-(8)(3)$

$A=84-24$

$A=60\ ft^2$

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

$A=b^2$

where

b is the length side of the square

so

$A=(12)(8)-(4^2)$

$A=96-16$

$A=80\ cm^2$

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

so

$A=4(4)(5)+(4^2)$

$A=80+16$

$A=96\ m^2$

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

so

$A=(15^2)-(9^2)$

$A=225-81$

$A=144\ cm^2$

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

where

b is the base of triangle

h is the height of triangle

so

$A=\frac{1}{2}(6)(7)-(6)(2)$

$A=21-12$

$A=9\ m^2$

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

so

$A=\pi (7^2)-(3)(11)$

$A=(49\pi -33)\ in^2$

7 0

3 years ago

The dimensions (width and length) of room1 have been read into two variables: width1 and length1. the dimensions of room2 have b

Reika [66]

Hello :
a single expression is : A = xy +zk
x : width1 y : length1
z : width2 k : length2
A : the total area of the two rooms
note :
A1 = xy ........area of the room 1
A2 = zk .......area of the room 2

3 0

3 years ago

Show that the relation R consisting of all pairs(x, y)such that x and y are bit strings of length three or more that agree in th

Mice21 [21]

Answer:

See proof below

Step-by-step explanation:

An equivalence relation R satisfies

Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
Symmetry: For all x,y, if xRy then yRx.
Transitivity: For all x,y,z, If xRy and yRz then xRz.

Let's check these properties: Let x,y,z be bit strings of length three or more

The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.

If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.

If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

ehidna [41]

Answer: 1) 0.10

2) 0.60

3) 0.20

4) 0.10

Step-by-step explanation:

The total frequency is 20+120+40+20 = 200. This means they ran the experiment 200 times. The probability distribution is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):

$\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 200}&\underline{\text{Probability Distribution}}\\1&20&20\div200=&0.10\\2&120&120\div 200=&0.60\\3&40&40\div 200=&0.20\\4&20&20\div 200=&0.10\end{array}\right]$

6 0

3 years ago

Hi guys I need help please

Zanzabum

Answer:

saannn Wala Naman

Step-by-step explanation:

HAHAHAJAH BLABLA

THANKS SA POINTS

7 0

3 years ago

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