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3 years ago

What is the factor of 25a2+b2?

Mathematics

1 answer:

Neko [114]3 years ago

7 0

Answer:

(5a+b)⋅(5a−b)

Step-by-step explanation:

Changes made to your input should not affect the solution:

(1): "b2" was replaced by "b^2". 1 more similar replacement(s).

STEP

Equation at the end of step 1

52a2 - b2

STEP

Trying to factor as a Difference of Squares

2.1 Factoring: 25a2-b2

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 25 is the square of 5

Check : a2 is the square of a1

Check : b2 is the square of b1

Factorization is : (5a + b) • (5a - b)

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What does it mean for a fraction to be in simplest form?

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A fraction is in it's simplest form if both of the numerator and denominator are not divisible by a number.

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3 years ago

A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one comp

Morgarella [4.7K]

Answer:

$y = 11 - 10cos(0.1\pi t)$

Step-by-step explanation:

Suppose at t = 0 the person is 1m above the ground and going up

Knowing that the wheel completes 1 revolution every 20s and 1 revolution = 2π rad in angle, we can calculate the angular speed

2π / 20 = 0.1π rad/s

The height above ground would be the sum of the vertical distance from the ground to the bottom of the wheel and the vertical distance from the bottom of the wheel to the person, which is the wheel radius subtracted by the vertical distance of the person to the center of the wheel.

$y = h_g + d_b = h_g + R - d_c$ (1)

where $h_g = 1$ is vertical distance from the ground to the bottom of the wheel, $d_b$ is the vertical distance from the bottom of the wheel to the person, R = 10 is the wheel radius, $d_c$ is the vertical distance of the person to the center of the wheel.

So solve for $d_c$ in term of t, we just need to find the cosine of angle θ it has swept after time t and multiply it with R

$d_c = Rcos(\theta(t)) = Rcos(\omega t) = 10cos(0.1\pi t)$

Note that $d_c$ is negative when angle θ gets between π/2 (90 degrees) and 3π/2 (270 degrees) but that is expected since it would mean adding the vertical distance to the wheel radius.

Therefore, if we plug this into equation (1) then

$y = h_g + R - d_c = 1 + 10 - 10cos(0.1\pi t) = 11 - 10cos(0.1\pi t)$