Solve the given system. y = x 20 y = 6x

1 answer:

8 0

If you would like to solve the following system, you can calculate this using the following steps:

y = x - 20
y = 6 * x
_________
6 * x = x - 20
6 * x - x = -20
5 * x = -20 /5
x = -4

y = 6 * x = 6 * (-4) = -24

If you would like to solve the following system, you can calculate this using the following steps:

y = x + 20
y = 6 * x
_________
6 * x = x + 20
6 * x - x = 20
5 * x = 20 /5
x = 4

y = 6 * x = 6 * 4 = 24

You might be interested in

What is the center of the circle that has a diameter whose endpoints are (9, 7) and (-3, -5)?

emmasim [6.3K]

Answer: OPTION B.

Step-by-step explanation:

Given the endpoints of the diameter of the circle, we need to find the midpoint, which is the center of the circle.

The following formula is used to calculate the Midpoint:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Given the endpoints of the diameter of the circle:

$(9, 7)\ and\ (-3, -5)$

We can identify that:

$x_1=9\\x_2=-3\\y_1=7\\y_2=-5$

Then, we can substitute these values into the formula:

$M=(\frac{9+(-3)}{2},\frac{7+(-5)}{2})\\\\M=(3,1)$

Therefore, the center of the circle is: $(3,1)$

3 0

3 years ago

12) The length of one side of a triangle is 24 cm and the length of its hypotenuse is 40 cm. find the length of its third side.

ivanzaharov [21]

Answer:

Step-by-step explanation:

given the triangle is right triangle

c^2 = a^2 + b^2

40^2 = 24^2 + b^2

1600 = 576 + b^2

b^2 = 1024

b = 32

6 0

2 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$