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laiz [17]
4 years ago
6

Describe a situation where it is easier to use decimals than fractions and explain why

Mathematics
1 answer:
valentina_108 [34]4 years ago
7 0
In money if you have £6.25 it is easier to write it in decimals
If you are measuring something e.g. 10.3cm long no one writes it 10 3/10 cm
You might be interested in
What is 2/8 in simpilist form​
Tems11 [23]

Answer:

1/4

Step-by-step explanation:

4 0
3 years ago
What is the missing term?
klio [65]

Answer:

-6

Step-by-step explanation:

3 x -2 = -6

You know this because you can see the pattern, whatever the column box is its always multiplied by the number its going on both vertical and horizontal.

6 0
3 years ago
Give that the point (15,-11) is on the graph of a one-to-one function f(x), find a point on the graph of f^{-1}.
zvonat [6]

Given:

The point (15,-11) is on the graph of a one-to-one function f(x).

To find:

The a point on the graph of f^{-1}.

Solution:

We know that, if the points on the graph of a one to one function are given by (x,y), then the points on the graph of its inverse function are given by (y,x).

In the point (15,-11).

x-coordinate = 15

y-coordinate = -11

In the inverse function,

x-coordinate = -11

y-coordinate = 15

Therefore, the required point on the graph of f^{-1} is (-11,15).

5 0
3 years ago
Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of K, where k
STALIN [3.7K]

\bold{\huge{\underline{\pink{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

  • <u>Rectangle </u><u>1</u><u> </u><u> </u><u>has </u><u>length </u><u>x</u><u> </u><u>and </u><u>width </u><u>y</u>
  • <u>Rectangle</u><u> </u><u>2</u><u> </u><u>is </u><u>made </u><u>by </u><u>multiplying </u><u>each </u><u>dimensions </u><u>of </u><u>rectangle </u><u>1</u><u> </u><u>by </u><u>a </u><u>factor </u><u>of </u><u>k </u>
  • <u>Where</u><u>, </u><u>k </u><u>></u><u> </u><u>0</u><u> </u><u> </u>

<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

Yes, The rectangle 1 and rectangle 2 are similar .

<h3><u>According </u><u>to </u><u>the </u><u>similarity </u><u>theorem </u><u>:</u><u>-</u></h3>

  • If the ratio of length and breath of both the triangles are same then the given triangles are similar.

<u>Let's </u><u>Understand </u><u>the </u><u>above </u><u>theorem </u><u>:</u><u>-</u>

The dimensions of rectangle 1 are x and y

<u>Now</u><u>, </u>

  • Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .

Let assume the value of K be 5

<u>Therefore</u><u>, </u>

The dimensions of rectangle 2 are

\sf{ 5x \:and \:5y }

<u>Now</u><u>, </u><u> </u><u>The </u><u>ratios </u><u>of </u><u>dimensions </u><u>of </u><u>both </u><u>the </u><u>rectangle </u><u>:</u><u>-</u>

  • \bold{Rectangle 1 =  Rectangle 2}

\bold{\dfrac{ x }{y}}{\bold{ = }}{\bold{\dfrac{5x}{5y}}}

\bold{\blue{\dfrac{ x }{y}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{x}{y}}}}

<u>From </u><u>above</u><u>, </u>

We can conclude that the ratios of both the rectangles are same

Hence , Both the rectangles are similar

<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>Here</u><u>, </u>

We have to proof that, the

  • Perimeter of rectangle 2 = k(perimeter of rectangle 1 )

In the previous questions, we have assume the value of k = 5

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

Perimeter of rectangle 1

\sf{ = 2( length + Breath) }

\bold{\pink{= 2( x + y ) }}

Thus, The perimeter of rectangle 1

Perimeter of rectangle 2

\sf{ = 2( length + Breath) }

\sf{ = 2(5x + 5y) }

\sf{ =  2 × 5( x + y) }

\bold{\pink{= 10(x + y) }}

Thus, The perimeter of rectangle 2

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

  • Perimeter of rectangle 2 = k( perimeter of rectangle 1 )

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 2(x + y) = 10(x + y)}

\bold{\pink{2x + 2y = 5(2x + 2y) }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.

Hence, The perimeter of rectangle 2 is k times of rectangle 1

<h3><u>Answer 3 :</u></h3>

<u>Here</u><u>, </u>

We have to proof that ,

  • <u>The </u><u>area </u><u>of </u><u>rectangle </u><u>2</u><u> </u><u>is </u><u>k²</u><u> </u><u>times </u><u>of </u><u>the </u><u>area </u><u>of </u><u>rectangle </u><u>1</u><u>.</u>

<u>That </u><u>is</u><u>, </u>

  • Area of rectangle 1 = k²( Area of rectangle)

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

<u>Area </u><u>of </u><u>rectangle </u><u>1</u>

\sf{ = Length × Breath }

\sf{ = x × y }

\bold{\red{= xy }}

<u>Area </u><u>of </u><u>rectangle </u><u>2</u>

\sf{ = Length × Breath }

\sf{ = 5x × 5y }

\bold{\red{ = 25xy }}

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

  • Area of rectangle 1 = k²( Area of rectangle)

\sf{ xy = 25xy }

\bold{\red{xy = (5)²xy }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5

Hence, The Area of rectangle 2 is k times of rectangle 1 .

7 0
2 years ago
Reflect AB over the x-axis and rotate 90° counterclockwise about the origin.​
Zepler [3.9K]

Answer:

O A"(3, -5) B"(1, -3)

Step-by-step explanation:

Hope it helps.

6 0
3 years ago
Read 2 more answers
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