Factor the following polynomial by grouping it in pairs.<br><br>4xy + 6 - x

How to find the center of the equation (x-7)²+(y-10)² = 121

grandymaker [24]

Center is 7,10
Radius is 11

3 0

2 years ago

8\3as a mixed number

Ber [7]

2 2/3

hope i helped. (:

7 0

2 years ago

Read 2 more answers

Line AB contains points A(-6,1) and B(1,4).

Free_Kalibri [48]

To solve this problem you must apply the proccedure shown below:

1- You have that the equation of the line is:

$y=mx+b$

Where $m$ is the slope and $b$ is the y-intercept.

2- Based on the information given in the problem, the lines $AB$ and $CD$ are parallel, which means that both have the same slope. Therefore, you can calculate the slope of $AB$ :

$m=\frac{y2-y1}{x2-x1}$

$m=\frac{4-1}{1-(-6)}=\frac{3}{7}$

3- Use the coordinates of the point $D(7,2)$ to calculate the y-intercept:

$2=\frac{3}{7}(7)+b$

4. Solve for $b$ :

$b=2-3=-1$

5. The equation of the line $CD$ is:

$y=\frac{3}{7}x-1$

The answer is: $y=\frac{3}{7}x-1$

6 0

3 years ago

What is 300 out of 2,530 as a percent

patriot [66]

In order to get the answer of that, let's understand first that the question can first be presented as a fraction form. So by breaking down the question, we will be able to get the fraction form which is 300/2530. Now in order to get the percent form of that, we simply divide 300 and 2530. The answer to that will result in 0.11858. Now we know for a fact that in order to make a decimal a percentage, we just have to move the decimal point two times to the right. So the answer is 11.858%

4 0

3 years ago

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Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$