Perimeter of rectangle = length + length + width + width

To find the combinations, think of two numbers that each multiplied by 2 and added up to give 12 or 14

Rectangle with perimeter 12

Say we take length = 2 and width = 3
Multiply the length by 2 = 2 × 2 = 4
Multiply the width by 3 = 2 × 3 = 6
Then add the answers = 4 + 6 = 10
This doesn't give us perimeter of 12 so we can't have the combination of length = 2 and width = 3

Take length = 4 and width = 2
Perimeter = 4+4+2+2 = 12
This is the first combination we can have

Take length = 5 and width = 1
Perimeter = 5+5+1+1 = 12
This is the second combination we can have

The question doesn't specify whether or not we are limited to use only integers, but if it is, we can only have two combinations of length and width that give perimeter of 12

length = 4 and width = 2
length = 5 and width = 1
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Rectangle with perimeter of 14

Length = 4 and width = 3
Perimeter = 4+4+3+3 = 14

Length = 5 and width = 2
Perimeter = 5+5+2+2 = 14

Length = 6 and width = 1
Perimeter = 6+6+1+1 = 14

We can have 3 different combinations of length and width

3 0

3 years ago

The midpoint is (blank,blank)

sveta [45]

Answer:

The answer to your question is Midpoint = (0, 1)

Equation of the bisector y = 2x + 1

Step-by-step explanation:

Data

P (-4, 3)

Q (4, -1)

Process

1.- Find the midpoint

Xm = (-4 + 4)/2

Xm = 0/2

Xm = 0

Ym = (3 - 1)/2

Ym = 2/2

Ym = 1

Midpoint = (0, 1)

2.- Equation of the line

slope = (-1 - 3) / (4 + 4)

= -4/8

= -1/2

But the new line is perpendicular so the new slope is

m = 2

Equation of the new line

y - y1 = m(x - x1)

y - 1 = 2(x - 0)

y - 1 = 2x

y = 2x + 1

6 0

3 years ago

How do i solve this question, anyone help

Degger [83]

Answer:

a. $\sqrt{52}$ cm or 7.21cm

b. $\sqrt{170}$ cm or 13.04cm

c. x= $\sqrt{136}$ cm or 11.66cm, y= $\sqrt{191}$ cm or 13.82cm

Step-by-step explanation:

a. You have to find the length of the other two sides of the triangle using the information already given. The first side is 6cm and the other is 12-9=4cm. Because it's a right-angled triangle you can use pythagoras

$c^2=a^2+b^2\\x^2=6^2+4^2\\x^2=36+16\\x^2=52\\x=\sqrt{52}cm \\x=7.21cm$