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3 years ago

Please help me anyone one, please

Mathematics

1 answer:

Elden [556K]3 years ago

3 0

Answer:

Step-by-step explanation:

Be sure to insert π where appropriate!

If the diameter of the yo-yo is 4 in, then the radius is 2 in.

The area comes from A = πr² and the circumference from C = πd.

In this case the area is A = π(2 in)² = 4π in² or 12.57 in².

The circumference is C = 4π in or 12.57 in.

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If on the unit circle C, the distance from P(1, 0) to the point T ( 4/5 , 3/5 ) is x, determine the coordinates of the point at

AveGali [126]

Answer:

$(\frac{4}{5}, -\frac{3}{5} )$

Step-by-step explanation:

The point P(1,0) and T $(\frac{4}{5}, \frac{3}{5} )$ are on the unit circle C and the arc length from P to T is x.

Let us assume that point P - x i.e. the point obtained by moving clock wise direction along the circle from P is T'.

Because of the symmetry of the circle about the coordinate axes, the x-coordinate of point T' will be $\frac{4}{5}$ and the y-coordinate will be $-\frac{3}{5}$ .

So, coordinates of T' are $(\frac{4}{5}, -\frac{3}{5} )$ .

Here we must notice that point T' is the reflection point of T with respect to X-axis. (Answer)

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3 years ago

Find the complete factored form of -15a^6b^5 + 12ab^2

Alisiya [41]

The answer should be −15a^6b^5+12ab^2

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3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$