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2 years ago

A regular pentagon and an equilateral triangle have the same perimeter...........

Mathematics

1 answer:

kondor19780726 [428]2 years ago

6 0

Answer:

Option 4: 45 units is the correct answer

Step-by-step explanation:

Let $P_1$ be the perimeter of Pentagon and

Let P2 be the perimeter of Triangle

So,

$P_1 = 5(\frac{1}{2}x-1)\\P_2 = 3(x-5)$

In order to find the perimeter of each figure we have to put both the expressions equal so that the value of x can be found and then perimeter.

So,

$P_1 = P_2\\5(\frac{1}{2}x-1) = 3(x-5)\\\frac{5}{2}x-5 = 3x-15\\\frac{5}{2}x-3x = -15+5\\\frac{5x-6x}{2} = -10\\-\frac{x}{2} = -10\\-x = -10 * 2\\-x = -20\\x = 20$

Putting x = 20 in 2nd perimeter

$P_2 = 3(x-5) \\= 3(20-5)\\= 3(15)\\=45$

As both the perimeters are same, the perimeter of pentagon will also be 45 units.

Hence,

Option 4: 45 units is the correct answer

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Find the length of the line segment formed by the points (2, 3) and (2, 5)

SCORPION-xisa [38]

Answer:

2 units

Step-by-step explanation:

Given data

The points are (2, 3) and (2, 5)

x1= 2

y1= 3

x2= 2

y2= 5

substitute into the distance between two points formula

d=√((x_2-x_1)²+(y_2-y_1)²)

d=√((2-2)²+(5-3)²)

d=√(2)²)

d=√4

d= 2 units

Hence the length is 2 units

3 0

2 years ago

ABDE is a rectangle on coordinate axes, the sides of the rectangle are parallel to the axes,

Hitman42 [59]

The coordinates of B and D are the location of the vertices B and D in the rectangle

The coordinates of B and D are (55, 30) and (55, 14)

<h3>How to determine the coordinates of B and D?</h3>

From the figure of the rectangle, we have:

A = (25,30)

C = (40, 22)

Point B has the same y-coordinate as point A

So, we have:

B = (x, 30)

The x-coordinate is then calculated as:

Bx = 2Cx - Ax

This gives

Bx = 2*40 - 25

Bx = 55

So, the coordinates of B are:

B = (55, 30)

Point D has the same x-coordinate as point B

So, we have:

D = (55, y)

The y-coordinate is then calculated as:

Dy = Cy - (By - Cy)

This gives

Dy = 22 - (30 - 22)

Dy = 14

So, the coordinates of B are:

D = (55, 14)

Hence, the coordinates of B and D are (55, 30) and (55, 14)

Read more about rectangles at:

brainly.com/question/10737082

8 0

2 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$