Answer:
B. ![\displaystyle (0, 4)\:and\:(3, 1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%280%2C%204%29%5C%3Aand%5C%3A%283%2C%201%29)
Step-by-step explanation:
☑
☑
☑
☑
I am joyous to assist you anytime.
Answer:
(2,10) or x=2 y=10
Step-by-step explanation:
<em>1. Pick one of your equations and solve for a variable. I chose the first equation and solved for x.</em>
5x-2y=-10 (Move the -2y to the other side, you need to do the opposite so you add +2y to -10)
5x=2y-10 (Divide the 5 from the x)
x=2/5y-2
<em>2. Now take what you got for x and plug it into the x variable on the other equation.</em>
3(2/5y-2)+6y=66 (Multiply 3 by 2/5y and -2)
6/5y-6=6y=66 (Move the -6 to the other side and add 6/5y to 6y)
36/5y=72 (Since the number on the y is a fraction, you must do the opposite to the other side)
y=72/1 x 5/36 (Flip your fraction and multiply it by the 72)
y=10
<em>3. Now that you have one of the variables solved for, in order to get the other we must plug in what we have to the first equation.</em>
5x-2(10)=-10 (Multiple 2 by 10)
5x-20=-10 (Move -20 to the other side, since you do the opposite add +20 to the -10)
5x=10 ( Divide 10 by 5)
x= 2
<em>4. If needed, plug in the values of x and y to check your solution.</em>
Hope this could help! :)
I believe the answer to this problem is C.<span />
The measure of the angle <JKL is 56 degrees
<h3>Tangent secant theorem of a circle</h3>
The theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
Given the following parameters
m<IL = 112 degrees
Since the measure of the vertex is half the measure of its intercepted arc, hence;
<JKl = 1/2(m<IL)
Substitute the given parameters into the formula to have;
<JKl = 1/2(112)
<JKL = 56 degrees
Hence the measure of the angle <JKL is 56 degrees
Learn more on Tangent secant theorem here: brainly.com/question/26826991
#SPJ1
I would say B but I can't see the question very well.