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3 years ago

1. Which equation does not represent a

Mathematics

2 answers:

aivan3 [116]3 years ago

6 0

B) y=x

isn’t a linear equation because it has to be written like y=mx+b

Murrr4er [49]3 years ago

6 0

Hdjsjskskdlslsndndbdkskndd

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AD = 6x - 4 and AC = 4x - 3

salantis [7]

Answer:

CD= 2x-1

Step-by-step explanation:

AD = 6x-4

AC = 4x - 3

CD= AD- AC

CD= 6x-4 - (4x - 3)

CD= 6x - 4 - 4x +3

CD= 2x -1

Therefore, the value of CD= 2x-1

3 0

1 year ago

Carlos dropped the phone he had received for his birthday and broke the screen. He knows he has to pay for the replacement himse

kiruha [24]

Answer:

????????????????

Step-by-step explanation:

what's the question

7 0

3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

3 years ago

Emeka is 10 years 6 months old and his classmate Boyle is 9 years old find the ratio of their age

Ghella [55]

Answer:

Hi...

10+6/12=10.5

So...10.5/9=1.16

4 0

3 years ago

Find the common difference of the sequence shown.

Vinvika [58]

The sequence is n/6. (:

5 0

3 years ago

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