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mrs_skeptik [129]
3 years ago
8

How do i solve a system of equation by graphing

Mathematics
1 answer:
jeyben [28]3 years ago
6 0

Hey there!


To solve a system of equation by graphing, first make sure both your equations are solved for y so that you can graph it (make sure it is in a form like: y=mx+b or y=ax^2+bx+c, etc).


Then, graph the equations on the graph, and the point where all the lines intersect would be your answer which should look like: (x,y)


However, if you have linear equations and the lines are parallel to each other, then there is no solution. If the lines graph to be the same line, then your answer would be all real solutions.

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Write an equation for the line that is parallel to the given line and passes through the given point.
Tresset [83]

Answer:

ll

Step-by-step explanation:

ll

3 0
2 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
Please help!
ss7ja [257]

Answer:

4 is the answer

8 0
2 years ago
Read 2 more answers
A nutritionist wants to know how the taste of artificial sweeteners impacts food choices of diabetic patients. From which group
kvasek [131]

The nutritionist should collect data from all diabetic patients to obtain accurate results.

<h3>What is the ideal population for this study?</h3>

This study aims at analyzing the food choices of diabetic patients. Because of this, the population studied should be diabetic patients including different types such as gestational, type I and II.

Moreover, because this affects to all diabetic patients, they all should be included even if they do not usually eat foods with arficial sweeteners.

Based on this, the best population is all diabetic patients.

Learn more about diabetic in: brainly.com/question/14823945

#SPJ1

6 0
2 years ago
Read 2 more answers
Look and solve plz herhenh
vodomira [7]

Answer:

s = 33

Step-by-step explanation:

99=3s

(99/3) = s

s=33

5 0
3 years ago
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