Which equation can be used to find one of the missing side lengths in the triangle?

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En

ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

7 0

3 years ago

Select the correct answer.<br> What is the solution to the equation?

Ierofanga [76]

Answer:

B

Step-by-step explanation:

$\sqrt{x}$ + 6 = x ( subtract 6 from both sides )

$\sqrt{x}$ = x - 6 ( square both sides )

x = (x - 6)² ← expand using FOIL

x = x² - 12x + 36 ( subtract x from both sides )

0 = x² - 13x + 36 , that is

x² - 13x + 36 = 0 ← in standard form

(x - 4)(x - 9) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x - 9 = 0 ⇒ x = 9

As a check

Substitute these values into the equation and if both sides are equal then they are the solutions.

x = 4

left side = $\sqrt{4}$ + 6 = 2 + 6 = 8

right side = x = 4

Since 8 ≠ 4 then x = 4 is an extraneous solution

x = 9

left side = $\sqrt{9}$ + 6 = 3 + 6 = 9

right side = x = 9

Thus the solution is x = 9 → B

8 0

3 years ago

Solve 2(1 – x) > 2x<br> this is my question for algebra

Rudik [331]

To solve this, you need to isolate/get the variable "x" by itself in the inequality:

2(1 - x) > 2x Divide 2 on both sides

$\frac{2(1-x)}{2} >\frac{2x}{2}$

1 - x > x Add x on both sides to get "x" on one side of the inequality

1 - x + x > x + x

1 > 2x Divide 2 on both sides to get "x" by itself

$\frac{1}{2} >x$ or $x$ (x is any number less than 1/2)

[Another way you could've solved it]

2(1 - x) > 2x Distribute 2 into (1 - x)

(2)1 + (2)(-x) > 2x

2 - 2x > 2x Add 2x on both sides

2 - 2x + 2x > 2x + 2x

2 > 4x Divide 4 on both sides to get "x" by itself

$\frac{2}{4} >\frac{4x}{4}$

$\frac{1}{2} >x$

6 0

3 years ago

Consider the quadratic function: y = 2x^2 + 7x + 10

arsen [322]

Answer:

y-intercept: 10

concavity: function opens up

min/max: min

Step-by-step explanation:

1.) The definition of a y-intercept is what the resulting value of a function is when x is equal to 0.

Therefore, if the function's equation is given, to find y-intercept simply plug in 0 for the x-values:

$y = 2x^2+7x+10 = 2(0)^2 + 7(0)+10 = 2(0) + 0 + 10 = 0 + 10 = 10$

y intercept ( f(0) )= 10

2.) In order to find concavity (whether a function opens up or down) of a quadratic function, you can simply find the sign associated with the x^2 value. Since 2x^2 is positive, the concavity is positive. This is basically possible, since it is identifying any reflections affecting the y-values / horizontal reflections.

3.) In order to find whether a quadratic function has a maximum or minimum, you can use the concavity of the function. The idea is that if the function opens downwards, the vertex would be at the very top, resulting in a maximum. If a function was open upwards, the vertex would be at the very bottom, meaning there is a minimum. Like the concavity, if the value associated with x^2 is positive, there is a minimum. If it is negative, there is a maximum. Since 2x^2 is positive, the function has a minimum.

7 0

3 years ago

Which of the following is NOT true? Question 19 options: (3x2y5)3=27x6y15 (1000x12y20z−3)0=1 52‾‾√3=523 42×43= 165

bixtya [17]

Answer:

Step-by-step explanation:

Lets try to simplify each to check which one is incorrect so it will be NOT true

first one is :

distribute the exponent so :

$(3x^2y^5)^3 =3^3x^{2*3}y^{5*3}$