The solution of such a system is the ordered pair that is a solution to both equations. To solve a system of linear equations graphically we graph both equations in the same coordinate system. The solution to the system will be in the point where the two lines intersect.

Step-by-step explanation:

7 0

3 years ago

Please answer this, I really need help

zhuklara [117]

Answer:

????? I am only in first grade

Step-by-step explanation:

???????

5 0

3 years ago

Can you help me with these too?<br> its for a geometry final practice<br> plisss

GrogVix [38]

The answer is D) They do not intersect and are not coplanar.

Hope this helps ;)

4 0

3 years ago

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THIS QUESTION IS EASY, I JUST NEED EXPLANATION.

chubhunter [2.5K]

It would help to have the question. But I'll guess

$\sqrt[4]{x^{\frac 3 5}}{\sqrt{x^2}} = ((x^{\frac 3 5})^{\frac 1 4}) }{ x} = x^{\frac 3{20}} x = x \sqrt[20]{x^3}$

2nd choice.

I forgot the explanation because I was so glad to have puzzled out the question.

The fourth root is the same as the 1/4 th power.

If we know x is positive $\sqrt{x^2}=|x|=x$

$\sqrt[4]{x^{\frac 3 5}}{\sqrt{x^2}}$

$= ((x^{\frac 3 5})^{\frac 1 4}) }{ x}$

We multiply the powers: $(a^b)^c=a^{bc}$

$= x^{\frac 3{20}} x$

The one-twentieth power is the twentieth root. We bring the additional factor of x to the front.

$= x \sqrt[20]{x^3}$

6 0

4 years ago

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