A circle and a square can intersect at how many points

1 answer:

3 0

If the circle has the same center as the diagonals of a square and the radius of the circle is smaller than 1/2 the diagonal of the square but larger than 1/2 the length of the side of a square, then there are 8 points of intersection -- 2 at each corner of the square.

If the radius of the circle is smaller than 1/2 the side length of the square and the center is as described above, there are no points of intersection.

If the circle is located outside the square it can have 1 tangent point or 2 intersection points depending on the location conditions of the circle in relation to the square.

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Find the area of this circle. Use 3 for a.A = 7r2=11 in[?] in?

liberstina [14]

SOLUTION

We want to find the area of the circle in the picture given.

We have been given the formula as

$\begin{gathered} A=\pi r^2 \\ We\text{ are told to take }\pi\text{ as 3} \\ \end{gathered}$

From the circle, the radius r = 11 in.

The area of the circle becomes

$\begin{gathered} A=\pi r^2 \\ A=\pi\times r^2 \\ A=3\times11^2 \\ A=3\times11\times11 \\ A=363in^2 \end{gathered}$

Hence the answer is 363 square-inches

7 0

11 months ago

Neil is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than u

Fynjy0 [20]

Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Step-by-step explanation:

Given:

Weight of a given sample (x) = 2.33 oz

Mean weight (μ) = 1.75 oz

Standard deviation (σ) = 0.22 oz

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

$z=\frac{x-\mu}{\sigma}$

Plug in the values and solve for 'z'. This gives,

$z=\frac{2.33-1.75}{0.22}=2.64$

So, the z-score of the distribution is 2.64.

Now, we need the probability $P(x\geq 2.33 )=P(z\geq 2.64)$ .

From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.

But, we need area more than the z-score value. So, the area is:

$P(z\geq 2.64)=1-0.9959=0.0041=0.41\%$