All categories

1 year ago

Fill in the blank with the correct rule

Mathematics

1 answer:

nlexa [21]1 year ago

8 0

Based on the given original and dilated points, you can notice that the dilation is:

(x,y) => (1/3 x , 1/3 y)

This can be verified with any point of the figure, for example:

C(6,-3) => C'(1/3 (6) , 1/3 (-3)) = C'(2 , -1)

Hence, the answer is:

(1/3 x , 1/3 y)

You might be interested in

How do you solve this?

puteri [66]

Answer:

Width: 7

Length: 14

Step-by-step explanation:

The area of a rectangle can be found my multiplying the length by the width. We do not know neither the length nor the width. However, we do know that the length is 7 feet longer than the width. Because we do not know the value of either side, let's let $x$ represent the width.

Since the length is 7 feet longer, we can represent the length by $x+7$ .

Now that we have our values for the length and the width we can multiply them together to find our area.

$(x)(x+7)=x^{2} +7x$

Now we have our equation, so we can address the changes made by the original question. The original question states that when 7 is added to both the length and width the area becomes 3 times larger. To do so simply increase each side by 7 by adding 7 to the original values.

$(x+7)(x+14)$

And multiply our area by 3.

$3(x^{2} +7x)$

set these equal to each other to find your new equation.

$(x+7)(x+14)=3(x^{2} +7x)$

Now you need to solve for $x$ . To do this first muliply $(x+7)$ and $(x+14)$ .

$(x+7)(x+14)=\\x^{2} +14x+7x+98=\\x^{2} +21x+98$

Then multiply $3(x^{2} +7x)$

$3(x^{2} +7x)=\\3x^{2} +21x$

Now you can begin solving for $x$ .

$x^{2} +21x+98=3x^{2} +21x$

Subtract $x^{2}$ from both sides.

$21x+98=2x^{2} +21x$

Subtract $21x$ from both sides.

$98=2x^{2}$

Divide by 2.

$49=x^{2}$

And finally take the sqaure root of both sides.

$\sqrt{x^{2} }=\sqrt{49}\\x=7$

Remember, because we are dealing with length, there cannot be negative. So while normally we would get both +7 and -7, in this case we only get +7.

Now that we have the value of $x$ , we can plug it into our original values.

For width we simply get 7.

For length we get 7+7=14

To check our answer we cna multiply 7 by 14 to get 98. Then we can add 7 to our length and width to get 14 and 21. Multiply these together and we get 294. 294 divided by 3 is 98, proving our answer correct.

7 0

4 years ago

A boat is marked up 20% on the original price. The original price was

pishuonlain [190]

Answer:

Original price was $50 x 0.20 = 10

original price $50 - 10 = $40 the sale price before sales tax

Step-by-step explanation:

hopefully it helps

3 0

2 years ago

Translate a algebraic expression .four less than four times a number. the translation is? type an expression using x as the vari

kicyunya [14]

4×√·²..................finish

3 0

4 years ago

The height of a triangle is half the length of its base. The area of the triangle is 12.25cm. Find the height

pentagon [3]

Answer: The height of the triangle is: " 3.5 cm " .
_______________________________________________________

Note: The formula/equation for the area, "A" , of a triangle is:

A = (1/2) * b * h ; or write as: A = (b * h) / 2 ;
_________________________________________________
in which: "A = area of the triangle" ;
"b = base length" ;
"h = "[perpendicular] height" ;
_________________________________________________
Given: h = (b/2) ;
A = 12.25 cm²
{Note: Let us assume that the given area was "12.25 cm² " .}.
_________________________________________________
We are to find the height, "h" ;

The formula for the Area, "A", is: A = (b * h) / 2 ;

Let us rearrange the formula ;
to isolate the "h" (height) on one side of the equation;

→ Multiply EACH side of the equation by "2" ; to eliminate the "fraction" ;

2*A = [ (b * h) / 2 ] * 2 ;

to get: " 2A = b * h " ;

↔ " b * h = 2A " ;

Divide EACH SIDE of the equation by "b" ; to isolate "h" on one side of the equation:

 → (b * h) / b = (2A) / b ;

to get:

 → h = 2A / b ;

Since "h = b/2" ; subtitute "b/2" for "h" ;

Plug in: "12.25 cm² " for "A" ;

→ b/2 = 2A/b ; → Note: " 2A/b = [2* (12.25 cm²) ] / b " ;

Note: " 2* (12.25 cm²) = 24.5 cm² ;

Rewrite as:

→ b/2 = (24.5 cm²) / b ;
_____________________________________
Cross-multiply: b*b = (24.5 cm²) *2 ;

to get: b² = 49 cm² ;

Take the "positive square root" of each side of the equation" ;
to isolate "b" on one side of the equation ; & to solve for "b" ;

→ +√(b²) = +√(49 cm²) ;

→ b = 7 cm ;

Now, we want to solve for "h" (the height) :
_________________________________________________________
→ h = b / 2 = 7 cm / 2 = 3.5 cm ;
_________________________________________________________
Answer: The height of the triangle is: " 3.5 cm " .
_________________________________________________________

6 0

3 years ago

The data sets show the years of the coins in two collections. Derek's collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

KATRIN_1 [288]

Answer:

Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

(1950+1952+ 1908+1902+1955+1954+1901+1910)/8

=1929

absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

from the mean of absolute deviation gives the MAD of the data i.e.

(21+23+21+27+26+25+28+`9)/8

23.75

:arrange the given data to get the range and median

1901 1902 1908 1910 1950 1952 1954 1955

The minimum value is: 1901

Maximum value is: 1955

Range is: Maximum value-minimum value

Range=1955-1901

Range= 54

median is (1910+1950)/2

1930

the lower set of data=

1901 1902 1908 1910

first quartile becomes

1902+1908)/2

Q1=1905

and upper set of data is:

1950 1952 1954 1955

we find the median of the upper quartile or third quartile is:

1952+1954)/2=1953

Q3-Q1=1953-1905=

IQR=48

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

1929+1935+ 1928+ 1930+ 1925+ 1932+1933+1920)/8

1929

absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

MAD=0+6+1+1+4+3+4+9/8

28/8

3.5

arrange the data in ascending order we get:

1920 1925 1928 1929 1930 1932 1933 1935

Minimum value= 1920

Maximum value= 1935

Range= 15 ( 1935-1920=15 )

The median is between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920 1925 1928 1929

the first quartile or upper quartile is

1925+1928/2

1926.5

and the upper set of data is:

1930 1932 1933 1935

We have

1932+1933)/2

1932.5

IQR is calculated as:

Q3-Q1

7 0

4 years ago