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mote1985 [20]
3 years ago
14

Is (-1,2) a solution of y=-2x-1

Mathematics
2 answers:
Klio2033 [76]3 years ago
7 0
A solution of a function is where the function hits the x-axis, so if y is any number that is not equal to 0, there is no possible way that point can be a solution. 

Hope this helps!
chubhunter [2.5K]3 years ago
4 0
The solution is: y=1.
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A parallelogram has verticies of (0,0) (0,6) (4,4) (4, -2). Which transformation maps the parallelogram onto itself?
gulaghasi [49]

The transformation that would map the parallelogram onto itself is the rotation of 180 degrees about the point (2, 2)

<h3>What is a transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

Rigid transformation is the transformation that does not change the shape or size of a figure. Examples of rigid transformations are <em>translation, reflection and rotation</em>.

The transformation that would map the parallelogram onto itself is the rotation of 180 degrees about the point (2, 2)

Find out more on transformation at: brainly.com/question/4289712

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5 0
1 year ago
What is the area of this triangle A=bh/2 <br> A:17m2<br> B:30m2<br> C:60m2<br> D:120m2
adelina 88 [10]
Base = 5
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Area = (5*12)/2 = 30 m²
5 0
3 years ago
How to do this problem <br> 3x-2=6
sergiy2304 [10]

Answer:

x = 2.667

Step-by-step explanation:

so first you wanna isolate the x by adding 2 to both sides

3x - 2 = 6 ---> 3x = 8

then divide both sides by 3 to get the x alone

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7 0
2 years ago
Read 2 more answers
6. Write equations of the following scenarios in ANY form
Bumek [7]

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answer is B

Step-by-step explanation:

7 0
3 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
3 years ago
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