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3 years ago

9

How many solution does an equation have when you isolate the variable and it equals a constant

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

Step-by-step explanation:

it depends. if (for example) y=x2 then there are an infinite amount of answers. if there is an equation like: If (variable X)= (variable Y) + 5 and if X=5, what is Y? then there is only one answer. check an algebra book, it can give you a more detailed answer.

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

kondaur [170]

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, $x_1=-8, y_1=-6, x_2=-4, y_2=-14$

$$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)$

$$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)$

$$P(x, y) =(-6, -10)$

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here, $x_1=-6, y_1=-10, x_2=-8, y_2=-6$

$r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}$

$r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}$

$r=\sqrt{(-2)^{2}+(4)^{2}}$

$r=\sqrt{20}$ units

The standard form of the equation of a circle is

$(x-a)^{2}+(y-b)^{2}=r^{2}$ , where (a, b) are center and r is the radius.

Here, center = (–6, –10) and $r=\sqrt{20}$

$(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}$

$(x+6)^{2}+(y+10)^{2}=20$

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

4 0

3 years ago

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

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

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

2 years ago

The graph of g(x) is a translation of y = 3/x

Crank

Answer:

d

Step-by-step explanation:

6 0

3 years ago

Please help with this algebra question

zhenek [66]

We have
-2x-3y <12
For this case what you should do is to graph the following straight line:
-2x-3y = 12
Then, the solution will be the shaded region that satisfies the following inequality:
-2x-3y <12
In the shaded region you will find all the solution points.
Answer:
See attached image

7 0

3 years ago

QUESTION 3 What is the area of the parallelogram?

lara31 [8.8K]

Answer:

C. 275.88‬ m squared

24.2 * 11.4 = 275.88‬ m squared

6 0

3 years ago