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3 years ago

Point I is on line segment HJ. Given HI = x, IJ = 2x + 9, and H J = 4x,

Mathematics

1 answer:

lutik1710 [3]3 years ago

5 0

:
length is 199
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first you do this then that

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3 years ago

1. What is the area of a circle with a diameter of 12.6 in.?

Rudik [331]

Answer:

Part 1) $A=124.63\ in^{2}$

Part 2) Option b

Part 3) As n increases, ns get closer to $2\pi r$

Part 4) Option c $16\pi\ ft^{2}$

Part 5) Option b. $12.25\pi\ m^{2}$

Step-by-step explanation:

Part 1) What is the area of a circle with a diameter of 12.6 in.?

we know that

the area of a circle is equal to

$A=\pi r^{2}$

we have

$r=12.6/2=6.3\ in$ -----> the radius is half the diameter

substitute the values

$A=(3.14)(6.3^{2})=124.63\ in^{2}$

Part 2) Which explanation can be used to derive the formula for the circumference of a circle?

First find the relationship of the circumference to its diameter by finding that the length of the diameter wraps around the length of the circumference approximately π times.

Use this relationship to write an equation showing the ratio of circumference to diameter equaling π

$\frac{C}{D}=\pi$

Rearrange the equation to solve for the circumference

$C=\pi D$

Substitute the diameter for 2 times the radius

$D=2r$

$C=2\pi r$

Part 3) we know that

If n increases

then

the product ns get closer to the circumference of the circle

the circumference of a circle is equal to $C=2\pi r$

therefore

As n increases, ns get closer to $2\pi r$

Part 4) What is the area of a circle whose radius is 4 ft?

we know that

the area of a circle is equal to

$A=\pi r^{2}$

we have

$r=4\ ft$

substitute the values

$A=(\pi)(4^{2})=16\pi\ ft^{2}$

Part 5) The circumference of a circle is 7π m.

What is the area of the circle?

we know that

The circumference of a circle is equal to

$C=2\pi r$

we have

$C=7\pi\ m$

substitute and solve for r

$7\pi=2\pi r$

$r=3.5\ m$

Find the area of the circle

the area of a circle is equal to

$A=\pi r^{2}$

substitute

the area of a circle is equal to

$A=\pi (3.5^{2})=12.25\pi\ m^{2}$

3 0

3 years ago