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3 years ago

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A sphere has a diameter of 8 cm. Which statements about the sphere are true? Check all that apply. The sphere has a radius of 4

cm. The sphere has a radius of 16 cm. The diameter’s length is twice the length of the radius. The radius’s length is twice the length of the diameter. The volume of the sphere is StartFraction 2,048 Over 3 EndFraction pi centimeters cubed. The volume of the sphere is StartFraction 256 Over 3 EndFraction pi centimeters cubed.

2 answers:

enyata [817]3 years ago

3 0

Answer:

1,2,4

Step-by-step explanation:

aev [14]3 years ago

3 0

Answer: The sphere has a radius of 4 cm.

The diameter’s length is twice the length of the radius

The volume of the sphere is StartFraction 256 Over 3 EndFraction pi centimeters cubed.

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The vertices of a triangle are A(4, 1), B(9, 1), C(2, 5). What is the area of this triangle.

kiruha [24]

Hey there!

$\large\boxed{10u^2}$

Formula for the area of a triangle is (1/2)bh.

Base of this triangle is 5, height is 4.

Image attached.

Plug values into the formula:

A = (1/2)(5)(4)

Simplify.

A = (1/2)(20)

A = 10

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Hope this helps!

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3 years ago

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

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$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

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$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

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The height of a model rocket is 4% of the height of the actual rocket. The model is 1.8 m high. How high is the actual rocket? (

Over [174]

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Luis es repartidor de leche el tiene 534 bolsas y las reparte entre cierto numero de tienda de modo que a cada le corresponde 28

OlgaM077 [116]

The first thing we must keep in mind is that there are 2 bags left over.
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3 years ago

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Hunter-Best [27]

32^2 feet so ya know

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