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3 years ago

Can Pie be Divided to make a Even Non-Decimal Number,if so what it it?

1 answer:

3 0

Yes, if you divide it by fractions of pi, then you would get even integers.

$\frac{\pi}{\pi/2} = 2 \\ \\ \frac{\pi}{\pi/4} = 4 \\ \\ \frac{\pi}{\pi/N} = N$

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There are 8 competitors in each ring for a tae kwon do tournament. If there are 96 competitors in the tournament, how many rings

Marysya12 [62]

All you need to do is divide 96 by 8 and you get 12 as your answer

5 0

2 years ago

Read 2 more answers

Find the value of x. please help me with my test

Lady bird [3.3K]

Step-by-step explanation:

180 -angled triangle,

180-56 = 124

124/2 = 62 (each corner of the right triangle)

then find the corner below

the rectangle has right angles on the right side

x = 90-62

= 28

8 0

2 years ago

This exercise involves the formula for the area of a circular sector. The area of a circle is 700 m2. Find the area of a sector

mezya [45]

Answer:

334.4 m²

Step-by-step explanation:

The formula for the area of a sector is given as:

1/2 × r² × θ

Where θ = Central angle

Area of a Circle = 700 m²

The formula for the area of a circle = πr²

r = Radius of a circle

r² = Area / π

r = √Area / π

r = √700/π

r = 14.927053304 m

Approximately, r = 14.93 m

Therefore, the area of the sector

= 1/2 × r² × θ

= 1/2 × 14.93² × 3 rad

= 334.35735 m²

Approximately, Area of the sector = 334.4 m²

4 0

2 years ago

Holly wants to make a frame for a painting. The painting is square and has an area of 225 square inches. a. Write an equation to

nalin [4]

Answer:

a) area = side²

225 in² = side²

side = √225 = 15 inches

b) 15 inches per side x 4 sides x $1.35 per inch = $81.00

Step-by-step explanation:

4 0

2 years ago

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Alik [6]

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

$(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0$

The general equation for the second-order polynomial is:

$x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}$

This condition must be observed for the case of a quadratic equation with equal roots:

$4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0$

$k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0$

$k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0$

$3\cdot k^{3} - 3\cdot k - 1 = 0$

7 0

3 years ago