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3 years ago

Figure ABCD is a kite. The area of ABCD is 48 square units. The length of line segment BD is 8 units. What is the length of AC?

Mathematics

2 answers:

Fed [463]3 years ago

8 0

A = d1 x d2 / 2

d1 = BD = 8 units

48 = 8 x AC / 2

48 x 2 = 8 AC

96 = 8 AC

AC = 96 : 8 = 12

Answer: D ) 12 units

bulgar [2K]3 years ago

5 0

The correct answer is:

D. 12 units

The length of AC is 12 units.

$|Huntrw6|$

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Olin [163]

Answer:

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then at (15,0) she spent all the money on (15) coffee and 0 orange juice

Step-by-step explanation:

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3 years ago

NEED HELP ASAP 3(x+2)=18

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Answer:

x=4

Step-by-step explanation:

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2 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

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2 years ago

Write the prime factorization of 500 using exponents.

vekshin1

2 x 2 x 5 x 5 x 5

2^2 x 5^3

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3 years ago

A company is designing a new cylindrical water bottle. The volume of the bottle will be 205 cm to the square root of 3 cm. The h

Umnica [9.8K]

Answer:

19 cm

Step-by-step explanation:

Given :

Volume, V = 205 to the √3 = 205^√3

Height, h = 8.9

radius, r

Volume of cylinder:

V = πr²h

205 to the square root of 3 10094.275

company is designing a new cylindrical water bottle. The volume of the bottle will be 205 cm to the square root of 3 cm. The height of the water bottle is 8.9 cm. What is the radius of the water bottle?

10094.275 = π * r² * 8.9

10094.275 = 3.14 * 8.9 * r²

10094.275 = 27.946r²

r² = 10094.275 / 27.946

r² = 361.20643

r = √361.20643

r = 19 cm

8 0

3 years ago