What the answer to y=2x+7 and y=4x-9 using substitution

Recall the equation for a circle with center (h, k) and radius r. At what point in the first quadrant does the line with equatio

lina2011 [118]

Answer:

The point of intersection is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (h, k) is the center and r is the radius.

Since our center is (0, 5), h = 0 and k = 5. The radius is 3. So, r = 3. Substitute:

$(x-0)^2+(y-5)^2=(3)^2$

Simplify:

$x^2+(y-5)^2=9$

At the point where the two equations intersect, its x-coordinate and y-coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for x. So:

$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

Square:

$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for x:

$\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$

Note that since we are looking for the point of intersection in QI, x should be positive. So, we can ignore the negative answer.

To find the y-coordinate, substitute the x-value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

6 0

3 years ago

The roots of the equation X square + 2 X + 4 = 2 are necessarily what

avanturin [10]

Answer:

-1±i

Step-by-step explanation:

Im assuming the question is asking what the roots of x^2+2x+4=2 are.

1. x^2+2x+4= 2 - subtract 2 on both sides to set the quadratic equation equal to zero.

2. x^2+2x+2 - now, since this equation cannot be factored the quadratic formula must be used: -b± $\frac{\sqrt{b^{2}-4ac}}{2a}$ (standard form: ax^2+bx+c)