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3 years ago

The lengths of the two diagonals of a rhombus are 6 cm and 8 cm. Find the length of its perimeter (in

Mathematics

1 answer:

adelina 88 [10]3 years ago

3 0

Answer:

The perimeter of the rhombus is 20 cm

Step-by-step explanation:

Perimeter of a Rhombus

Given the lengths of the two diagonals of a rhombus, let's call them a and b, the perimeter of the rhombus is given by:

$P=2\sqrt{a^2+b^2}$

The values of the diagonals provided by the question are a=6 cm, b=8 cm, thus the perimeter is:

$P=2\sqrt{6^2+8^2}$

$P=2\sqrt{36+64}$

$P=2\sqrt{100}=2*10$

$P=20\ cm$

The perimeter of the rhombus is 20 cm

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12. If DH = HE, MBG = (9x – 20) and MGC = (5x + 28), find mAB.

svetoff [14.1K]

Answer:

176degrees

Step-by-step explanation:

Find the diagram attached. From the diagram;

arcGC = = arc BG

5x + 28 = 9x - 20

5x - 9x = -20 28

-4x = -48

x = -48/-4

x = 12

Get <AB

Since <AB = <BC

<BC = arcGC +arc BG

<AB = arcGC+ arc BG

<AB = 5x + 28 + 9x - 20

<AB = 14x + 8

<AB = 14(12) + 8

<AB = 168 + 8

<AB = 176degrees

Hence the arc AB is 176degrees

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3 years ago

What are the value(s) of x for tan(x) = 0?

Nesterboy [21]

Answer:

4.the answer is option four

Step-by-step explanation:

tan(360)=0

tan(180)=0

tan(0)=0

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3 years ago

Ben has 3 times as many guppies as goldfish. If he has a total of 20 fish, how may guppies does he have?

olchik [2.2K]

15 guppies 5 goldfish ,

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3 years ago

Question is in the picture A. 3 B. 7 C. 5 D. 6

marishachu [46]

I think the answer is 7. Hope this helps!

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4 years ago

Q 2 PLEASE HELP ME FIGURE THIS OUT

suter [353]

Answer: IV, positive, $\frac{\pi} {6}$ , - sec $\frac{\pi} {6}$ , $\frac{2\sqrt{3}}{3}$

Step-by-step explanation:

a) Look at the Unit Circle to see that $\frac{11\pi} {6}$ = 330°, which is located in Quadrant IV.

b) The coordinate (cos θ, sin θ) for $\frac{11\pi} {6}$ is: $(\frac{\sqrt{3}} {2},\frac{-1}{2})$

sec = $\frac{1}{cos}$ = $\frac{2}{\sqrt{3}}$ which is positive

c) Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the given angle $\frac{11\pi} {6}$ from 2π: $\frac{12\pi} {6}$ - $\frac{11\pi} {6}$ = $\frac{\pi} {6}$

d) the reference angle is below the x-axis so the given angle is equal to the negative of the reference angle: - sec $\frac{\pi} {6}$ .

e) sec $\frac{11\pi} {6}$ = $\frac{2}{\sqrt{3}}$ = $\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{3}$

***************************************************************************************

Answer: $\frac{18\pi}{11}$ , IV, $\frac{4\pi} {11}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{22\pi}{11}$

$\frac{-26\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{-4\pi}{11}$

$\frac{-4\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{18\pi}{11}$

Convert the radians into degrees to see which Quadrant it is in by setting up the proportion and cross multiplying:

$\frac{\pi}{180}$ = $\frac{18\pi}{11x}$

π(11x) = (180)18π

x = $\frac{180(18\pi}{11\pi}$

x = 295° which lies in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{18\pi} {11}$ from 2π: $\frac{22\pi} {11}$ - $\frac{18\pi} {11}$ = $\frac{4\pi} {11}$

***************************************************************************************

Answer: $\frac{5\pi}{3}$ , IV, $\frac{4\pi} {11}$ , $\frac{\pi} {3}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{6\pi}{3}$

$\frac{-13\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-7\pi}{3}$

$\frac{-7\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-\pi}{3}$

$\frac{-\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{5\pi}{3}$

This is on the Unit Circle at 300°, which is located in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{5\pi} {3}$ from 2π: $\frac{6\pi} {3}$ - $\frac{5\pi} {3}$ = $\frac{\pi} {3}$

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4 years ago