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kondor19780726 [428]

3 years ago

12

What are the coordinates of point BBB on \overline{AC} AC start overline, A, C, end overline such that the ratio of ABABA, B to

BCBCB, C is 2:32:32, colon, 3?
A) 2 1/5, 1/5
B) 2 1/4, 1/4
C) 3 3/4, 3/4
D) 3 3/5, 3/5

2 answers:

nikdorinn [45]3 years ago

5 0

Answer:

Stepb-by-step explanation:

Nady [450]3 years ago

3 0

I think it’s gonna be C

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Which statement regarding the function y=sin(x) is true?

REY [17]

<h2>Answer:</h2>

c. Reflection over either the x-axis or y-axis will change the graph

<h2>Step-by-step explanation:</h2><h3>a. Reflection over the y-axis will not change the graph since sine is an even function.</h3>

This is false because $y=sin(x)$ is an odd function, not an even one. This means that $sin(-x)=-sin(x)$ , and a reflection over the y-axis will change the graph.

<h3>b. Sin(x)=sin(-x)</h3>

This is false because we said that $sin(-x)=-sin(x)$

<h3>c. Reflection over either the x-axis or y-axis will change the graph</h3>

This is true. Since $sin(x)$ is an odd function, then reflection over either the x-axis or y-axis will change the graph as we said in a. So, for $f(x)$ :

<u>REFLEXION IN THE X-AXIS:</u>

$h(x)=-f(x)$

<u>REFLEXION IN THE Y-AXIS:</u>

$h(x)=f(-x)$

<h3>d. Sin(x)=-sin(x)</h3>

False by the same explanation as b.

6 0

4 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

or

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

or

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒ $x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

8 0

4 years ago

Read 2 more answers

3/4 •12/25<br> How to simplify

Firlakuza [10]

Answer:

36/100

Step-by-step explanation:

3/4 * 12/25 = 36/100

multiply the 2 numerators together, then multiple the 2 denominators together.

hope this helps

3 0

3 years ago

Read 2 more answers

What is the range of the function shown in the graph below?

navik [9.2K]

Answer:

A

Step-by-step explanation:

5 0

3 years ago

How to factor 9x^4-6x^2y^4+y^89x4−6x2y4+y8

Virty [35]

Answer:

$(3x^2-y^4)(3x^2-y^4)$

Step-by-step explanation:

We'll assume the correct expression to be factored is

$9x^4-6x^2y^4+y^8$

One must try to find out if the expression is a perfect square. To test it, we'll take the square root of the first and the last term. If they are exact, we'll procceed with the next step

$\sqrt{9x^4}=3x^2$

$\sqrt{y^8}=y^4$

Since both are exact, we'll test if the middle term is twice the product of both square roots:

$2(3x^2)(y^4)=6x^2y^4$

We confirm the middle term equals the above expression. All tests confirm the original expression is

$\left ( 3x^2-y^4 \right )^2$

The required factoring is

$(3x^2-y^4)(3x^2-y^4)$

8 0

3 years ago