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baherus [9]
3 years ago
8

Danessa needs to compare the area of one large circle with a diameter of 8 to the total area of 2 smaller circles with a diamete

r one-half that of the large circle. Which statements about the areas are true? Check all that apply.
1.The radius of the large circle is 4.
2.The radii of the small circles are each 2.
3.The radii of the small circles are each 4.
4.The area of one small circle will be one-half of the area of the large circle.
5.The total area of the two small circles will equal that of the large circle.
6.The total area of the two small circles will be one-half of the area of the large circle.
Mathematics
2 answers:
BaLLatris [955]3 years ago
4 0

Answer:

The answer is: 1, 2, and 6

Step-by-step explanation:

Nat2105 [25]3 years ago
3 0

Answer with Step-by-step explanation:

Diameter of larger circle=8

Diameter of smaller circles=4

We know that radius is half times the diamketer

and Area=πr² where r is the radius

Radius of larger circle=4

Radius of smaller circles=2

Area of larger circle=π×4²

                                =16π

Area of two smaller circles=2×π×2²

                                           = 8π=one-half the area of larger circle

Area of one small circle=π×2²

                                       =4π=one fourth the area of larger circle

Hence, the correct options are:

1.The radius of the large circle is 4.

2.The radii of the small circles are each 2.

6.The total area of the two small circles will be one-half of the area of the large circle.

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S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
3 0
3 years ago
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igomit [66]

Answer:

The original price was $312.50

Step-by-step explanation:

If the vacuum was marked down by 20%, that means it was sold for 80% of the full price. It was sold for $250, so 80% of the original price is $250.

Let's set up an equation.

80/100 of x = 250

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Divide both sides by 0.8.

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2 years ago
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Umnica [9.8K]

Answer:

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Step-by-step explanation:

Distance formula:

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Coordinates:

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