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4 years ago

Least conmon denominator of 5/8, 1/2, 5/6

Mathematics

1 answer:

grin007 [14]4 years ago

5 0

the least common denominator of the given fractions is 24.

hope this helped.

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A designer enlarged both the length and the width of a rectangular carpet by 60 percent. The new carpet was too large so the des

Snowcat [4.5K]

x= length, y=width

Both the length and width were increased by 60% so now they are:

1.6x and 1.6y

Then they were both reduced by 25% so now they are 75%(100-25) of their previous size so they are:

0.75(1.6x) and 0.75(1.6y)

Their area would be length * width so it would be:

0.75(1.6x) * 0.75(1.6y)

Simplified this would be:

1.2x * 1.2y

Simplified more it would be :

1.44xy which is also 144% of xy which is a 44% increase of the original area (xy)

So the answer is:

44%

5 0

3 years ago

What is pi times 34 times pi

zhannawk [14.2K]

The answer is

Pi times 34 times pi= 335.6

7 0

3 years ago

Read 2 more answers

Special triangles 2 back

AleksAgata [21]

Answer:

Step-by-step explanation:

10. From the given figure,

$\frac{18}{y}=sin60^{0}$

$y=\frac{36}{\sqrt{3}}$

11. From the given figure,

$\frac{12}{y}=sin60^{0}$

$y=\frac{12{\times}2}{\sqrt{3}}=\frac{24}{\sqrt{3}}$

and $\frac{x}{y}=cos60^{0}$

$x=\frac{24}{\sqrt{3}{\times}2}=\frac{12}{\sqrt{3}}$

12. The sides of an equilateral triangle has length=10 inches, therefore AB=BC=AC=10

Also, AD is the height of the given triangle,then BD=DC=5

From triangle ADC,

$(AC)^{2}=(AD)^{2}+(DC)^{2}$

$100=(AD)^{2}+25$

$75=(AD)^{2}$

$AD=\sqrt{75}inches$

Hence, height of the triangle is $\sqrt{75}inches$

13. Height of the ramp from the ground is 8ft.

therefore, let BC=x

$\frac{8}{BC}=tan30^{0}$

$\frac{8}{x}=\frac{1}{\sqrt{3}}$

$x=8\sqrt{3}$

14. Let BC=x and AB=y

then, $\frac{x}{18}=cos30^{0}$

$x=18{\times}\frac{\sqrt{3}}{2}=9\sqrt{3}$

and $\frac{y}{x}=tan30^{0}$

$y=9$

The perimeter of triangle is=Sum of all the sides=AB+BC+AC= $9\sqrt{3}+9+18= 9\sqrt{3}+27=9(1.73)+27=42.57 inches$

15. Let AB be the height of the tree, then

$\frac{AB}{8}=tan30^{0}$

$AB=8{\times}\frac{1}{\sqrt{3}}=\frac{8}{1.73}=4.62m$

16. According to question,

$\frac{h}{24}=sin30^{0}$

$h=\frac{24}{2}=12ft$

7 0

3 years ago

Read 2 more answers

What is the length of arc PQ? What is the area of the shaded region?

Liula [17]

Comment
I'm assuming you mean the minor arc. That's the part of the circumference that is in the shaded part.

Part One
The length of the arc is connected to the central angle which you are given as 155o
r = 9 in
The arc length = $\theta*r * 2\pi /360$
Since $\theta$ = 155 degrees

The arc length = 2*3.14 * 9*155 / 360
The arc length = 24.3 inches

The sector area = (central angle/360) * pi * r^2 .
The sector area = (155/360) * 3.14 * 9^2
The sector area = 109.5 in^2

Answers
Arc length = 24.3 <<<<<<< answer
Arc area = 109.5 <<<<<<<<answer

7 0

3 years ago

From the diagram below, if

Veseljchak [2.6K]

The possible values of x in the diagram are x = 25 and x = -1

<h3>How to determine the value of x?</h3>

The given parameters are:

∠DAB = x² - 24x
∠DAC = 25

Both angles are congruent and adjacent angles.

So, we have:

∠DAB = ∠DAC

Substitute known values

x² - 24x = 25

Subtract 25 from both sides

x² - 24x - 25 = 0

Expand

x² + x - 25x - 25 = 0

Factorize

x(x + 1) - 25(x + 1) = 0

Factor out x + 1

(x - 25)(x + 1) = 0

Expand

x - 25 = 0 or x + 1 = 0

Solve for x

x = 25 or x = -1

Hence, the possible values of x are x = 25 and x = -1